模板（持续更新）

发表于 更新于 本文字数： 44k 阅读时长 ≈ 40 分钟

左手栏有目录。

出现的宏：

1
2
3
4
5
6
7
8
#define rep(i, l, r) for(int i = (l); i <= (r); i++)
#define per(i, r, l) for(int i = (r); i >= (l); i--)
#define mem(a, b) memset(a, b, sizeof a)
#define For(i, l, r) for(int i = (l), i##e = (r); i < i##e; i++)
#define pb push_back

using ll = long long;
using lf = double;

有用的模板

Fast IO

非负整数版

函数
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
int read() {
    const int M = 1e6;
    static streambuf* in = cin.rdbuf();
    #define gc (p1 == p2 && (p2 = (p1 = buf) + in -> sgetn(buf, M), p1 == p2) ? -1 : *p1++)
    static char buf[M], *p1, *p2;
    int c, r = 0;
    while((c = gc) < 48);
    while(c > 47) r = r * 10 + (c & 15), c = gc;
    return r;
}
void wrt(int x) {
    static streambuf* out = cout.rdbuf();
    #define pc out -> sputc
    static char c[11]; int sz = 0;
    do c[++sz] = x % 10, x /= 10; while(x);
    while(sz) pc(c[sz--] + 48);
    pc(10);
}
预处理
1
cin.tie()->sync_with_stdio(0);

整数版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
int read() {
    const int M = 1e6;
    static streambuf* in = cin.rdbuf();
    #define gc (p1 == p2 && (p2 = (p1 = buf) + in -> sgetn(buf, M), p1 == p2) ? -1 : *p1++)
    static char buf[M], *p1, *p2;
    int c, r = 0, f = 1;
    while((c = gc) < 48) if(c == 45) f = -1;
    while(c > 47) r = r * 10 + (c & 15), c = gc;
    return r * f;
}
void wrt(int x) {
    static streambuf* out = cout.rdbuf();
    #define pc out -> sputc
    static char c[11]; int sz = 0;
    if(x < 0) pc(45), x = -x;
    do c[++sz] = x % 10, x /= 10; while(x);
    while(sz) pc(c[sz--] + 48);
    pc(10);
}

注：输出 long long 时 wrt 函数中的 c 数组大小要开到 \(20\)。

AtCoder-modint

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using uint = unsigned;
using ull = unsigned ll;

constexpr ll safe_mod(ll x, ll m) {
    return x %= m, x < 0 ? x + m : x;
}
constexpr ll pow_mod_constexpr(ll x, ll n, int m) {
    if(m == 1) return 0;
    uint _m = m;
    ull r = 1, _x = safe_mod(x, m);
    for(; n; n >>= 1, _x = _x * _x % _m)
    if(n & 1) r = r * _x % m;
    return r;
}
constexpr bool is_prime_constexpr(int n) {
    if(n <= 1) return false;
    if(n == 2 || n == 7 || n == 61) return true;
    if(n % 2 == 0) return false;
    ll d = n - 1;
    while(~d & 1) d >>= 1;
    for(ll a : {2, 7, 61}) {
        ll t = d, y = pow_mod_constexpr(a, t, n);
        while(t != n - 1 && y != 1 && y != n - 1)
            y = y * y % n, t <<= 1;
        if(y != n - 1 && t % 2 == 0) return false;
    }
    return true;
}
constexpr pair<ll, ll> inv_gcd(ll a, ll b) {
    a = safe_mod(a, b);
    if(a == 0) return {b, 0};
    ll s = b, t = a, m0 = 0, m1 = 1;
    while(t) {
        ll u = s / t;
        s -= t * u, m0 -= m1 * u;
        ll tmp = s;
        s = t, t = tmp, tmp = m0, m0 = m1, m1 = tmp;
    }
    if(m0 < 0) m0 += b / s;
    return {s, m0};
}
struct barrett {
    uint m; ull im;
    barrett(uint m) :m(m), im(~0ull / m + 1) {}
    uint mul(uint a, uint b) const {
        ull z = (ull)a * b;
        ull x = (__uint128_t)z * im >> 64;
        uint v = z - x * m;
        return m <= v ? v + m : v;
    }
};
template<int m> struct static_modint {
    using mint = static_modint;
  public:
    static mint raw(int v) {
        mint x;
        return x._v = v, x;
    }
    static_modint() : _v(0) {}
    template<class T> static_modint(T v) {
        ll x = v % m;
        _v = x < 0 ? x + m : x;
    }
    uint val() const { return _v; }
    mint& operator++() {
        if(++_v == m) _v = 0;
        return *this;
    }
    mint& operator--() {
        if(!_v--) _v = m - 1;
        return *this;
    }
    mint operator++(int) {
        mint res = *this;
        ++*this;
        return res;
    }
    mint operator--(int) {
        mint res = *this;
        --*this;
        return res;
    }
    mint& operator+=(const mint& rhs) {
        _v += rhs._v;
        if(_v >= m) _v -= m;
        return *this;
    }
    mint& operator-=(const mint& rhs) {
        _v -= rhs._v;
        if(_v >= m) _v += m;
        return *this;
    }
    mint& operator*=(const mint& rhs) {
        ull z = _v;
        z *= rhs._v, _v = z % m;
        return *this;
    }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }

    mint pow(ll n) const {
        assert(0 <= n);
        mint x = *this, r = 1;
        for(; n; n >>= 1, x *= x) if(n & 1) r *= x;
        return r;
    }
    mint inv() const {
        if(prime) {
            assert(_v);
            return pow(m - 2);
        } else {
            auto eg = inv_gcd(_v, m);
            assert(eg.first == 1);
            return eg.second;
        }
    }

    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs) += rhs;
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs) -= rhs;
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs) *= rhs;
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs) /= rhs;
    }
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs._v == rhs._v;
    }
    friend bool operator!=(const mint& lhs, const mint& rhs) {
        return lhs._v != rhs._v;
    }

  private:
    uint _v;
    static constexpr bool prime = is_prime_constexpr(m);
};

template<int id> struct dynamic_modint {
    using mint = dynamic_modint;

  public:
    static void set_mod(int m) {
        assert(1 <= m), bt = barrett(m);
    }
    static mint raw(int v) {
        mint x;
        return x._v = v, x;
    }

    dynamic_modint() : _v(0) {}
    template<class T> dynamic_modint(T v) {
        ll x = v % (int)bt.m;
        _v = x < 0 ? x + bt.m : x;
    }

    uint val() const { return _v; }

    mint& operator++() {
        if(++_v == bt.m) _v = 0;
        return *this;
    }
    mint& operator--() {
        if(!_v--) _v = bt.m - 1;
        return *this;
    }
    mint operator++(int) {
        mint res = *this;
        ++*this;
        return res;
    }
    mint operator--(int) {
        mint res = *this;
        --*this;
        return res;
    }

    mint& operator+=(const mint& rhs) {
        _v += rhs._v;
        if(_v >= bt.m) _v -= bt.m;
        return *this;
    }
    mint& operator-=(const mint& rhs) {
        _v += bt.m - rhs._v;
        if(_v >= bt.m) _v -= bt.m;
        return *this;
    }
    mint& operator*=(const mint& rhs) {
        _v = bt.mul(_v, rhs._v);
        return *this;
    }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }

    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }

    mint pow(ll n) const {
        assert(0 <= n);
        mint x = *this, r = 1;
        for(; n; n >>= 1, x *= x) if(n & 1) r *= x;
        return r;
    }
    mint inv() const {
        auto eg = inv_gcd(_v, bt.m);
        assert(eg.first == 1);
        return eg.second;
    }

    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs) += rhs;
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs) -= rhs;
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs) *= rhs;
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs) /= rhs;
    }
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs._v == rhs._v;
    }
    friend bool operator!=(const mint& lhs, const mint& rhs) {
        return lhs._v != rhs._v;
    }

  private:
    uint _v;
    static barrett bt;
};
template<int id> barrett dynamic_modint<id>::bt = 998244353;

打比赛专用模板

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using uint = unsigned;
using ull = uint64_t;

constexpr ll safe_mod(ll x, ll m) { return x %= m, x < 0 ? x + m : x; }
constexpr ll pow_mod_constexpr(ll x, ll n, int m) {
    if(m == 1) return 0;
    uint _m = m; ull r = 1, _x = safe_mod(x, m);
    for(; n; n >>= 1, _x = _x * _x % _m) if(n & 1) r = r * _x % m;
    return r;
}
constexpr bool is_prime_constexpr(int n) {
    if(n <= 1) return false;
    if(n == 2 || n == 7 || n == 61) return true;
    if(n % 2 == 0) return false;
    ll d = n - 1; while(~d & 1) d /= 2;
    for(ll a : {2, 7, 61}) {
        ll t = d, y = pow_mod_constexpr(a, t, n);
        while(t != n - 1 && y != 1 && y != n - 1) y = y * y % n, t <<= 1;
        if(y != n - 1 && t % 2 == 0) return false;
    }
    return true;
}
constexpr pair<ll, ll> inv_gcd(ll a, ll b) {
    a = safe_mod(a, b);
    if(a == 0) return {b, 0};
    ll s = b, t = a, m0 = 0, m1 = 1;
    while(t) {
        ll u = s / t; s -= t * u, m0 -= m1 * u;
        ll tmp = s; s = t, t = tmp, tmp = m0, m0 = m1, m1 = tmp;
    }
    if(m0 < 0) m0 += b / s;
    return {s, m0};
}
struct barrett {
    uint m; ull im;
    barrett(uint m) :m(m), im(~0ull / m + 1) {}
    uint mul(uint a, uint b) const {
        ull z = (ull)a * b, x = (__uint128_t)z * im >> 64; uint v = z - x * m;
        return m <= v ? v + m : v;
    }
};
template<int m> struct static_modint {
    using mint = static_modint;
  public:
    static mint raw(int v) { mint x; return x._v = v, x; }
    static_modint() :_v(0) {}
    template<class T> static_modint(T v) { ll x = v % m; _v = x < 0 ? x + m : x; }
    uint val()const { return _v; }
    mint& operator++() { if(++_v == m) _v = 0; return *this; }
    mint& operator--() { if(!_v--) _v = m - 1; return *this; }
    mint operator++(int) { mint res = *this; ++*this; return res; }
    mint operator--(int) { mint res = *this; --*this; return res; }
    mint& operator+=(const mint& rhs) { _v += rhs._v; if(_v >= m) _v -= m; return *this; }
    mint& operator-=(const mint& rhs) { _v -= rhs._v; if(_v >= m) _v += m; return *this; }
    mint& operator*=(const mint& rhs) { ull z = _v; z *= rhs._v, _v = z % m; return *this; }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
    mint operator+()const { return *this; }
    mint operator-()const { return mint() - *this; }
    mint pow(ll n)const { assert(0 <= n); mint x = *this, r = 1; for(; n; n >>= 1, x *= x) if(n & 1) r *= x; return r; }
    mint inv() const{ if(prime) { assert(_v); return pow(m - 2); } else { auto eg = inv_gcd(_v, m); assert(eg.first == 1); return eg.second; } }
    friend mint operator+(const mint& lhs, const mint& rhs) { return mint(lhs) += rhs; }
    friend mint operator-(const mint& lhs, const mint& rhs) { return mint(lhs) -= rhs; }
    friend mint operator*(const mint& lhs, const mint& rhs) { return mint(lhs) *= rhs; }
    friend mint operator/(const mint& lhs, const mint& rhs) { return mint(lhs) /= rhs; }
    friend bool operator==(const mint& lhs, const mint& rhs) { return lhs._v == rhs._v; }
    friend bool operator!=(const mint& lhs, const mint& rhs) { return lhs._v != rhs._v; }
  private:
    uint _v;
    static constexpr bool prime = is_prime_constexpr(m);
};
struct dynamic_modint {
    using mint = dynamic_modint;
  public:
    static void set_mod(int m) { assert(1 <= m), bt = barrett(m); }
    static mint raw(int v) { mint x; return x._v = v, x; }
    dynamic_modint() :_v(0) {}
    template<class T> dynamic_modint(T v) { ll x = v % (int)bt.m; _v = x < 0 ? x + bt.m : x; }
    uint val()const { return _v; }
    mint& operator++() { if(++_v == bt.m) _v = 0; return *this; }
    mint& operator--() { if(!_v--) _v = bt.m - 1; return *this; }
    mint operator++(int) { mint res = *this; ++*this; return res; }
    mint operator--(int) { mint res = *this; --*this; return res; }
    mint& operator+=(const mint& rhs) { _v += rhs._v; if(_v >= bt.m) _v -= bt.m; return *this; }
    mint& operator-=(const mint& rhs) { _v += bt.m - rhs._v; if(_v >= bt.m) _v -= bt.m; return *this; }
    mint& operator*=(const mint& rhs) { _v = bt.mul(_v, rhs._v); return *this; }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
    mint operator+()const { return *this; }
    mint operator-()const { return mint() - *this; }
    mint pow(ll n)const { assert(0 <= n); mint x = *this, r = 1; for(; n; n >>= 1, x *= x) if(n & 1) r *= x; return r; }
    mint inv()const { auto eg = inv_gcd(_v, bt.m); assert(eg.first == 1); return eg.second; }
    friend mint operator+(const mint& lhs, const mint& rhs) { return mint(lhs) += rhs; }
    friend mint operator-(const mint& lhs, const mint& rhs) { return mint(lhs) -= rhs; }
    friend mint operator*(const mint& lhs, const mint& rhs) { return mint(lhs) *= rhs; }
    friend mint operator/(const mint& lhs, const mint& rhs) { return mint(lhs) /= rhs; }
    friend bool operator==(const mint& lhs, const mint& rhs) { return lhs._v == rhs._v; }
    friend bool operator!=(const mint& lhs, const mint& rhs) { return lhs._v != rhs._v; }
  private:
    uint _v;
    static barrett bt;
};
barrett dynamic_modint::bt = 998244353;
#define rep(i, l, r) for(int i = (l); i <= (r); i++)
#define per(i, r, l) for(int i = (r); i >= (l); i--)
#define mem(a, b) memset(a, b, sizeof a)
#define For(i, l, r) for(int i = (l), i##e = (r); i < i##e; i++)
#define pb push_back
#define eb emplace_back
#define fi first 
#define se second
#define all(x) (x).begin(), (x).end()
#define SZ(x) int((x).size())
#define mid ((l + r) >> 1)
#define lc (o << 1)
#define rc (o << 1 | 1)
#define lch l, mid, lc
#define rch mid + 1, r, rc
#define lb lower_bound
#define ub upper_bound
#define bs binary_search

template<class T> T& cmin(T& a, const T& b) { if(b < a) a = b; return a; }
template<class T> T& cmax(T& a, const T& b) { if(a < b) a = b; return a; }

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

void solve() {
    
}
int main() {
    cin.tie(0)->sync_with_stdio(0);
    int T;
    for(cin >> T; T--; solve());
}

随机数

函数

1
2
3
4
5
6
7
8
9
10
11
12
using ull = unsigned ll;
ull gen(ull x) {
    const ull k = 0x9ddfea08eb382d69ULL;
    x *= k, x ^= x >> 47;
    x *= k, x ^= x >> 47;
    x *= k, x ^= x >> 47;
    return x * k;
}
int rnd() {
    static ull s = 2;
    return (s += gen(s)) & INT_MAX;
}

大模数取模

函数

1
2
3
ll mul(ll a, ll b, ll p) {
    return (a * b - ll((long double)a / p * b + 0.5) * p + p) % p;
}

bash 对拍

1
2
3
4
5
6
7
while true; do
    ./gen > in
    ./a < in > 1
    ./b < in > 2
    diff 1 2
    if [ $? -ne 0 ] ; then break; fi
done

简易计算器

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
#include <bits/stdc++.h>
#define rep(i, l, r) for(int i = (l); i <= (r); i++)
#define per(i, r, l) for(int i = (r); i >= (l); i--)
#define mem(a, b) memset(a, b, sizeof a)

using namespace std;
using ll = long long;

ll mul(ll a, ll b, ll p) {
    return (a * b - ll((long double)a / p * b + 0.5) * p + p) % p;
}
ll Pow(ll a, ll n, ll p, ll r = 1) {
    for(; n; n >>= 1, a = mul(a, a, p))
    if(n & 1) r = mul(r, a, p);
    return r;
}

namespace Pollard_Rho {
    int chk(ll n) {
        for(ll a : {2, 3, 7, 61, 24251}) {
            if(n == a) return 1;
            if(Pow(a, n - 1, n)  1) return 0;
            ll k = n - 1, t;
            while(~k & 1) k >>= 1;
            if((t = Pow(a, k, n)) == 1) continue;
            while(t != 1 && t != n - 1) t = mul(t, t, n);
            if(t != n - 1) return 0;
        }
        return 1;
    }
    ll f(ll x, ll c, ll p) { return (mul(x, x, p) + c) % p; }
    ll PR(ll n) {
        ll a = 0, b = 0, c = rand() % (n - 1) + 1, v = 1, g;
        for(int k = 1; ; k <<= 1, a = b, v = 1) {
            rep(i, 1, k) {
                b = f(b, c, n), v = mul(v, abs(a - b), n);
                if(!(i & 127) || i == k) {
                    g = __gcd(v, n);
                    if(g > 1) return g;
                }
            }
        }
    }
    ll ans[100]; int cnt;
    void solve(ll n) {
        if(chk(n)) return void(ans[++cnt] = n);
        ll d; do d = PR(n); while(d == n);
        solve(d), solve(n / d);
    }
    void work() {
        ll n;
        while(cin >> n) {
            cnt = 0;
            if(chk(n)) { puts("Prime"); continue; }
            solve(n), sort(ans + 1, ans + cnt + 1);
            int t = 0;
            rep(i, 1, cnt) {
                if(ans[i] != ans[i - 1]) cout << ans[i];
                t++; if(ans[i] != ans[i + 1]) {
                    if(t > 1) cout << '^' << t;
                    putchar(32), t = 0;
                }
            } putchar(10);
        }
    }
};
namespace Inv {
    void exgcd(ll a, ll b, ll& d, ll& x, ll& y) {
        if(b) exgcd(b, a % b, d, y, x), y -= a / b * x;
        else d = a, x = 1, y = 0;
    }
    void inv(ll a, ll p) {
        ll d, x, y; exgcd(a, p, d, x, y);
        if(d > 1) puts("Non-existent!");
        else cout << (x % p + p) % p;
    }
    void work() {
        ll a, p;
        while(cin >> a >> p) inv(a, p);
    }
};
namespace Prime {
    void work() {
        ll n;
        while(cin >> n) {
            while(!Pollard_Rho::chk(n)) n++;
            cout << n << endl;
        }
    }
};
namespace Cipolla {
    ll n, p, II;
    struct cmp {
        ll r, i;
        cmp operator *(const cmp& b) {
            return {(r * b.r + i * b.i % p * II) % p, (r * b.i + i * b.r) % p };
        }
    } U = { 1, 0 };
    int pow1(ll a, int n, ll r = 1) {
        for(; n; n >>= 1, a = a * a % p)
        if(n & 1) r = r * a % p;
        return r;
    }
    cmp pow2(cmp a, int n, cmp r = U) {
        for(; n; n >>= 1, a = a * a)
        if(n & 1) r = r * a;
        return r;
    }
    void work() {
        while(cin >> n >> p) {
            if(!n) { puts("0"); continue; }
            if(p == 2) { cout << n << endl; continue; }
            if(pow1(n, p / 2) != 1) { puts("Non-existent!"); continue; }
            ll a;
            do a = rand() % p, II = (a * a - n + p) % p; while(!a || pow1(II, p / 2) == 1);
            int x1 = pow2({a, 1}, p / 2 + 1).r, x2 = p - x1;
            if(x1 > x2) swap(x1, x2);
            printf("%d %d\n", x1, x2);
        }
    }
}; 
int main() {
    srand(time(0));
    puts("Press 1 for integer factorization.");
    puts("Press 2 to calculate the modular multiplicative inverse of a number.");
    puts("Press 3 to find the first prime number greater than or equal to a number.");
    puts("Press 4 to calculate the modular square root of a number.");
    int op;
    while(1) {
        cin >> op;
        if(op == 1) Pollard_Rho::work();
        else if(op == 2) Inv::work();
        else if(op == 3) Prime::work();
        else if(op == 4) Cipolla::work();
        else puts("Illegal input! Please re-enter your option.");
    }
}

数学

NTT

普通版

定义

1
int lim = 1, rev[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
ll Pow(ll a, int n, ll r = 1) {
    for(; n; n >>= 1, a = a * a % P)
    if(n & 1) r = r * a % P;
    return r;
}
void NTT(int a[], bool t) {
    For(i, 0, lim) if(rev[i] < i) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1) {
        ll wn = Pow(3, P / 2 / i);
        for(int j = 0; j < lim; j += i << 1) {
            ll w = 1;
            For(k, j, j + i) {
                int x = a[k], y = a[k + i] * w % P;
                a[k] = (x + y) % P, a[k + i] = (x - y + P) % P, w = w * wn % P;
            }
        }
    }
    if(t) return;
    reverse(a + 1, a + lim);
    ll inv = Pow(lim, P - 2);
    For(i, 0, lim) a[i] = a[i] * inv % P;
}

modint 版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
void NTT(mint a[], bool t) {
    For(i, 0, lim) if(rev[i] < i) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1) {
        mint wn = mint(3).pow(P / 2 / i);
        for(int j = 0; j < lim; j += i << 1) {
            mint w = 1;
            For(k, j, j + i) {
                mint x = a[k], y = a[k + i] * w;
                a[k] = x + y, a[k + i] = x - y, w *= wn;
            }
        }
    }
    if(t) return;
    reverse(a + 1, a + lim);
    mint inv = mint(lim).inv();
    For(i, 0, lim) a[i] *= inv;
}

预处理

1
2
while(lim <= n + m) lim <<= 1;
For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 1);

较快版

定义

1
int lim = 1, rev[N], w[N];

modint 版

1
2
mint w[N];
int lim = 1, rev[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
int add(int a, int b) { return (a += b) < P ? a : a - P; }
int sub(int a, int b) { return (a -= b) < 0 ? a + P : a; }
ll Pow(ll a, int n, ll r = 1) {
    for(; n; n >>= 1, a = a * a % P)
    if(n & 1) r = r * a % P;
    return r;
}
void NTT(int a[], bool t) {
    For(i, 0, lim) if(rev[i] < i) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
    for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
        int x = a[k], y = (ll)a[k + i] * w[i + k - j] % P;
        a[k] = add(x, y), a[k + i] = sub(x, y);
    }
    if(t) return;
    reverse(a + 1, a + lim);
    ll inv = Pow(lim, P - 2);
    For(i, 0, lim) a[i] = a[i] * inv % P;
}

modint 版

1
2
3
4
5
6
7
8
9
10
11
12
void NTT(mint a[], bool t) {
    For(i, 0, lim) if(rev[i] < i) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
    for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
        mint x = a[k], y = a[k + i] * w[i + k - j];
        a[k] = x + y, a[k + i] = x - y;
    }
    if(t) return;
    reverse(a + 1, a + lim);
    mint inv = mint(lim).inv();
    For(i, 0, lim) a[i] *= inv;
}

预处理

1
2
3
4
5
6
while(lim <= n + m) lim <<= 1;
For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 0);
for(int i = 1; i < lim; i <<= 1) {
    ll wn = Pow(3, P / 2 / i), p = 1;
    For(j, 0, i) w[i + j] = p, p = p * wn % P;
}

modint 版

1
2
3
4
5
6
while(lim <= n + m) lim <<= 1;
For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 0);
for(int i = 1; i < lim; i <<= 1) {
    mint wn = mint(3).pow(P / 2 / i), p = 1;
    For(j, 0, i) w[i + j] = p, p *= wn;
}

任意模数 NTT / MTT

定义

1
2
3
4
5
6
7
using cmp = complex<lf>;
const lf PI = acos(-1);
const cmp I(0, 1);

int n, m, P;
int M, lim = 1, rev[N];
cmp w[N], a0[N], a1[N], b0[N], b1[N];

modint 版

1
2
3
4
5
6
using cmp = complex<lf>;
const lf PI = acos(-1);

int n, m, P;
int M, lim = 1, rev[N];
cmp w[N], a0[N], a1[N], b0[N], b1[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
void FFT(cmp a[], bool t) {
    For(i, 0, lim) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
    for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
        cmp x = a[k], y = a[k + i] * w[i + k - j];
        a[k] = x + y, a[k + i] = x - y;
    }
    if(t) return;
    reverse(a + 1, a + lim);
    lf inv = 1. / lim;
    For(i, 0, lim) a[i] *= inv;
}
void FFT2(cmp a[], cmp b[]) {
    For(i, 0, lim) a[i] += b[i] * I;
    FFT(a, 0);
    For(i, 0, lim) b[i] = conj(a[i ? lim - i : 0]);
    For(i, 0, lim) {
        cmp x = a[i], y = b[i];
        a[i] = (y + x) * 0.5, b[i] = (y - x) * 0.5i;
    }
}
ll num(cmp x) { return M * ll(real(x) + 0.5) % P + ll(imag(x) + 0.5); }

modint 版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
void FFT(cmp a[], bool t) {
    For(i, 0, lim) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
    for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
        cmp &x = a[k], y = a[k + i] * w[i + k - j];
        a[k + i] = x - y, x += y;
    }
    if(t) return;
    reverse(a + 1, a + lim);
    lf inv = 1. / lim;
    For(i, 0, lim) a[i] *= inv;
}
void FFT2(cmp a[], cmp b[]) {
    For(i, 0, lim) a[i] += b[i] * 1i;
    FFT(a, 0);
    For(i, 0, lim) b[i] = conj(a[i ? lim - i : 0]);
    For(i, 0, lim) {
        cmp x = a[i], y = b[i];
        a[i] = (y + x) * 0.5, b[i] = (y - x) * 0.5i;
    }
}
mint num(cmp x) { return M * (mint)ll(real(x) + 0.5) % P + (mint)ll(imag(x) + 0.5); }

预处理

1
2
3
4
5
6
7
M = sqrt(P);
rep(i, 0, n) a0[i] = A[i] / M, a1[i] = A[i] % M;
rep(i, 0, m) b0[i] = B[i] / M, b1[i] = B[i] % M;
while(lim <= n + m) lim <<= 1;
For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 0);
for(int i = 1; i < lim; i <<= 1) For(j, 0, i)
    w[i + j] = cmp(cos(PI / i * j), sin(PI / i * j));

使用

1
2
3
4
5
6
7
FFT2(a0, a1), FFT2(b0, b1);
For(i, 0, lim) {
    cmp t = a0[i] + I * a1[i];
    b0[i] *= t, b1[i] *= t;
}
FFT(b0, 1), FFT(b1, 1);
rep(i, 0, n + m) C[i] = (M * num(b0[i]) + num(b1[i])) % P;

modint 版

1
2
3
4
5
6
7
FFT2(a0, a1), FFT2(b0, b1);
For(i, 0, lim) {
    cmp t = a0[i] + I * a1[i];
    b0[i] *= t, b1[i] *= t;
}
FFT(b0, 1), FFT(b1, 1);
rep(i, 0, n + m) C[i] = M * num(b0[i]) + num(b1[i]);

FWT

函数

1
2
3
4
5
6
7
8
9
10
11
12
void FWT(ll a[]) {
    For(i, 0, n) For(S, 0, 1 << n) if(S >> i & 1) {
        ll& x = a[S ^ 1 << i], y = a[S];
        a[S] = x - y, x += y;
    }
}
void IFWT(ll a[]) {
    For(i, 0, n) For(S, 0, 1 << n) if(S >> i & 1) {
        ll& x = a[S ^ 1 << i], y = a[S];
        a[S] = x - y >> 1, x = x + y >> 1;
    }
}

多项式求逆 / inv

定义

1
int n, lim, rev[N], w[N];

modint 版

1
2
int n, lim, rev[N];
mint w[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
int add(int a, int b) { return (a += b) >= P ? a - P : a; }
int sub(int a, int b) { return (a -= b) < 0 ? a + P : a; }
ll Pow(ll a, int n, ll r = 1) {
    for(; n; n >>= 1, a = a * a % P)
    if(n & 1) r = r * a % P;
    return r;
}
void NTT(int a[], bool t) {
    For(i, 0, lim) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
        for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
            int x = a[k], y = (ll)a[k + i] * w[i + k - j] % P;
            a[k] = add(x, y), a[k + i] = sub(x, y);
        }
    if(t) return;
    reverse(a + 1, a + lim);
    ll inv = Pow(lim, P - 2);
    For(i, 0, lim) a[i] = a[i] * inv % P;
}
void Inv(int n, int a[], int b[]) {
    for(int k = 1; k <= n; k <<= 1) {
        lim = k << 2;
        memcpy(c, a, lim << 1);
        For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 0);
        for(int i = 1; i < lim; i <<= 1) {
            ll wn = Pow(3, P / i / 2), p = 1;
            For(j, i, i << 1) w[j] = p, p = p * wn % P;
        }
        NTT(b, 1), NTT(c, 1);
        For(i, 0, lim) b[i] = (ll)b[i] * sub(2, (ll)c[i] * b[i] % P) % P;
        NTT(b, 0), memset(b + lim / 2, 0, lim << 1);
    }
}

modint 版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
void NTT(mint a[], bool t) {
    For(i, 0, lim) if(rev[i] < i) swap(a[i], a[rev[i]]);
    for(int i = 1; i < lim; i <<= 1)
    for(int j = 0; j < lim; j += i << 1) For(k, j, j + i) {
        mint x = a[k], y = a[k + i] * w[i + k - j];
        a[k] = x + y, a[k + i] = x - y;
    }
    if(t) return;
    reverse(a + 1, a + lim);
    mint inv = mint(lim).inv();
    For(i, 0, lim) a[i] *= inv;
}
void Inv(int n, mint a[], mint b[]) {
    for(int k = 1; k <= n; k <<= 1) {
        lim = k << 2;
        memcpy(c, a, lim << 1);
        For(i, 0, lim) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? lim >> 1 : 0);
        for(int i = 1; i < lim; i <<= 1) {
            mint wn = mint(3).pow(P / i / 2), p = 1;
            For(j, i, i << 1) w[j] = p, p = p * wn;
        }
        NTT(b, 1), NTT(c, 1);
        For(i, 0, lim) b[i] *= 2 - c[i] * b[i];
        NTT(b, 0), memset(b + lim / 2, 0, lim << 1);
    }
}

自然数等幂求和

中国剩余定理

扩展中国剩余定理 / exCRT

1
2
3
4
5
6
7
8
9
10
11
ll mul(ll a, ll b, ll p) {
    return (a * b - (ll)((long double)a / p * b + 0.5) * p + p) % p;
}
void exgcd(ll a, ll b, ll& d, ll& x, ll& y) {
    if(!b) { d = a, x = 1, y = 0; return; }
    exgcd(b, a % b, d, y, x), y -= a / b * x;
}
void exCRT(ll& b1, ll& m1, ll b2, ll m2) {
    ll d, k1, k2; exgcd(m1, m2, d, k1, k2), m2 /= d;
    b1 = (b1 + mul(mul(k1 % m2, (b2 - b1) / d % m2, m2), m1, m1 * m2)) % (m1 *= m2);
}

杜教筛

Min-25 筛 / min25

定义

1
ll f1[N], f2[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
ll min25(ll n) {
    int lim = sqrt(n);
    rep(i, 1, lim) f1[i] = i - 1, f2[i] = n / i - 1;
    rep(p, 2, lim) if(f1[p] != f1[p - 1]) {
        int w1 = lim / p;
        ll x = f1[p - 1], w3 = (ll)p * p, w2 = min((ll)lim, n / w3), d = n / p;
        rep(i, 1, w1) f2[i] -= f2[i * p] - x;
        rep(i, w1 + 1, w2) f2[i] -= f1[d / i] - x;
        per(i, lim, w3) f1[i] -= f1[i / p] - x;
    }
    return f2[1];
}

exBSGS

定义

1
map<int, int> mp;

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
int BSGS(ll pls, ll a, ll b, int p) {
    pls %= p, a %= p, b %= p; mp.clear();
    ll m = ceil(sqrt(p)), ls = 1, rs = 1;
    For(i, 0, m) mp[ls * b % p] = i, ls = ls * a % p;
    rep(i, 1, m) {
        rs = rs * ls % p;
        if(mp.count(rs * pls % p)) return i * m - mp[rs * pls % p];
    }
    return -1;
}
int exBSGS(int a, int b, int p) {
    a %= p, b %= p;
    int pls = 1, cnt = 0, g;
    while((g = __gcd(a, p)) > 1) {
        if(b == pls) return cnt;
        if(b % g) return -1;
        p /= g, b /= g, pls = pls * ll(a / g) % p, cnt++;
    }
    int ret = BSGS(pls, a, b, p);
    return ~ret ? ret + cnt : -1;
}

cipolla

定义

1
2
3
4
5
6
7
int n; ll II;
struct cmp {
    ll r, i;
    cmp operator *(const cmp& b) {
        return {(r * b.r + i * b.i % P * II) % P, (r * b.i + i * b.r) % P};
    }
} U = {1, 0};

modint 版

1
2
3
4
5
6
7
mint n, II;
struct cmp {
    mint r, i;
    cmp operator *(const cmp& b) {
        return {r * b.r + i * b.i * II, r * b.i + i * b.r};
    }
} U = {1, 0};

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
int pow1(ll a, int n, ll r = 1) {
    for(; n; n >>= 1, a = a * a % P)
    if(n & 1) r = r * a % P;
    return r;
}
cmp pow2(cmp a, int n, cmp r = U) {
    for(; n; n >>= 1, a = a * a)
    if(n & 1) r = r * a;
    return r;
}
int cipolla(int n) {
    if(!n) return 0;
    if(P == 2) return n;
    if(pow1(n, P / 2) != 1) return -1;
    ll a;
    do a = rand() % P, II = (a * a - n + P) % P; while(!a || pow1(II, P / 2) == 1);
    return pow2({a, 1}, P / 2 + 1).r;
}

modint 版

1
2
3
4
5
6
7
8
9
10
11
12
13
cmp Pow(cmp a, int n, cmp r = U) {
    for(; n; n >>= 1, a = a * a)
    if(n & 1) r = r * a;
    return r;
}
int cipolla(mint n) {
    if(n == 0) return 0;
    if(P == 2) return n.val();
    if(n.pow(P / 2) != 1) return -1;
    mint a;
    do a = rand(), II = a * a - n; while(a == 0 || II.pow(P / 2) == 1);
    return Pow({a, 1}, P / 2 + 1).r.val();
}

Miller Rabin & Pollard Rho

定义

1
vector<ll> as;

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
ll mul(ll a, ll b, ll p) {
    return (a * b - ll((long double)a / p * b + 0.5) * p + p) % p;
}
ll Pow(ll a, ll n, ll p, ll r = 1) {
    for(; n; n >>= 1, a = mul(a, a, p))
    if(n & 1) r = mul(r, a, p);
    return r;
}
int chk(ll n) {
    for(ll a : {2, 3, 7, 61, 24251}) {
        if(n == a) return 1;
        if(Pow(a, n - 1, n) != 1) return 0;
        ll k = n - 1, t;
        while(~k & 1) k >>= 1;
        if((t = Pow(a, k, n)) == 1) continue;
        while(t != 1 && t != n - 1) t = mul(t, t, n);
        if(t != n - 1) return 0;
    }
    return 1;
}
ll f(ll x, ll c, ll p) { return (mul(x, x, p) + c) % p; }
ll PR(ll n) {
    ll a = 0, b = 0, c = rand() % (n - 1) + 1, v = 1, g;
    for(int k = 1; ; k <<= 1, a = b, v = 1) {
        rep(i, 1, k) {
            b = f(b, c, n), v = mul(v, abs(a - b), n);
            if(!(i & 127) || i == k) {
                g = __gcd(v, n);
                if(g > 1) return g;
            }
        }
    }
}
void solve(ll n) {
    if(chk(n)) return as.pb(n);
    ll d; do d = PR(n); while(d == n);
    solve(d), solve(n / d);
}

数据结构

动态树 / LCT

普通版

定义

1
int c[N][2], f[N], r[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
bool id(int o) { return c[f[o]][1] == o; }
bool nrt(int o) { return c[f[o]][0] == o || c[f[o]][1] == o; }
void pu(int o) {
    
}
void rev(int o) {
    swap(c[o][0], c[o][1]), r[o] ^= 1;
}
void pd(int o) { if(r[o]) rev(c[o][0]), rev(c[o][1]), r[o] = 0; }
void rot(int o, int d) {
    int k = c[o][!d], &x = c[k][d];
    if(nrt(o)) c[f[o]][id(o)] = k;
    pu(x = f[c[o][!d] = x] = o), f[k] = f[o], pu(f[o] = k);
}
void dfs(int o) { if(nrt(o)) dfs(f[o]); pd(o); }
void splay(int o) {
    dfs(o);
    for(int fa; nrt(o); rot(f[o], !id(o)))
    if(nrt(fa = f[o])) rot(id(o) == id(fa) ? f[fa] : fa, !id(o));
}
void acc(int o) {
    for(int x = 0; o; o = f[x = o])
        splay(o), c[o][1] = x, pu(o);
}
void mkrt(int o) { acc(o), splay(o), rev(o); }
void link(int u, int v) {
    mkrt(u), acc(v), splay(v), pu(f[u] = v);
}
void cut(int u, int v) {
    mkrt(u), acc(v), splay(v), c[v][0] = f[u] = 0, pu(v);
}

维护子树 size 版

定义

1
int c[N][2], f[N], r[N], s[N], si[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
bool id(int o) { return c[f[o]][1] == o; }
bool nrt(int o) { return c[f[o]][0] == o || c[f[o]][1] == o; }
void pu(int o) {
    s[o] = s[c[o][0]] + s[c[o][1]] + si[o] + 1;
}
void rev(int o) {
    swap(c[o][0], c[o][1]), r[o] ^= 1;
}
void pd(int o) { if(r[o]) rev(c[o][0]), rev(c[o][1]), r[o] = 0; }
void rot(int o, int d) {
    int k = c[o][!d], &x = c[k][d];
    if(nrt(o)) c[f[o]][id(o)] = k;
    pu(x = f[c[o][!d] = x] = o), f[k] = f[o], pu(f[o] = k);
}
void dfs(int o) { if(nrt(o)) dfs(f[o]); pd(o); }
void splay(int o) {
    dfs(o);
    for(int fa; nrt(o); rot(f[o], !id(o)))
    if(nrt(fa = f[o])) rot(id(o) == id(fa) ? f[fa] : fa, !id(o));
}
void acc(int o) {
    for(int x = 0; o; o = f[x = o])
        splay(o), si[o] += s[c[o][1]] - s[x], c[o][1] = x, pu(o);
}
void mkrt(int o) { acc(o), splay(o), rev(o); }
void link(int u, int v) {
    mkrt(u), acc(v), splay(v), si[v] += s[u], pu(f[u] = v);
}
void cut(int u, int v) {
    mkrt(u), acc(v), splay(v), c[v][0] = f[u] = 0, pu(v);
}

pbds::tree

定义

1
2
3
4
5
6
7
8
9
10
11
12
#include <bits/extc++.h>

using namespace __gnu_pbds;
template<class T> using Set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class K, class V> using Map = tree<K, V, less<K>, rb_tree_tag, tree_order_statistics_node_update>;
#define ins(T, x) (T).insert(x)
#define era(T, x) (T).erase(x)
#define rk(T, x) ((T).order_of_key(x) + 1)
#define kth(T, x) (*(T).find_by_order(x - 1))
#define pre(T, x) (*--(T).lower_bound(x))
#define nxt(T, x) (*(T).upper_bound(x))
#define bld(T, l, r) (T).copy_from_range(l, r)

RBS 树

普通版

定义

1
int rt, sz, lc[N], rc[N], c[N], s[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
using ull = unsigned ll;
ull fn(ull x) {
    const ull k = 0x9ddfea08eb382d69ULL;
    x *= k, x ^= x >> 47;
    x *= k, x ^= x >> 47;
    x *= k, x ^= x >> 47;
    return x * k;
}
int rand(int mod) {
    static ull s = 2;
    return (s += fn(s)) % mod;
}
int pu(int o) { s[o] = s[lc[o]] + s[rc[o]] + 1; return o; }
void bld(int l, int r, int& o) {
    if(l > r) return;
    int mid = l + r >> 1;
    c[o = ++sz] = a[mid];
    bld(l, mid - 1, lc[o]), bld(mid + 1, r, rc[o]), pu(o); 
}
void spt(int o, int x, int& u, int& v) {
    if(!o) u = v = 0;
    else if(x < c[o]) spt(lc[v = o], x, u, lc[o]), pu(o);
    else spt(rc[u = o], x, rc[o], v), pu(o);
}
int mrg(int u, int v) {
    if(!u || !v) return u + v;
    if(rand(s[u] + s[v]) < s[u])
        return rc[u] = mrg(rc[u], v), pu(u);
    return lc[v] = mrg(u, lc[v]), pu(v);
}
void ins(int& o, int x) {
    if(rand(s[o] + 1) == 0)
        c[++sz] = x, spt(o, x, lc[sz], rc[sz]), pu(o = sz);
    else ins(x < c[o] ? lc[o] : rc[o], x), pu(o);
}
void rmv(int& o, int x) {
    int t1, t2, t3;
    spt(o, x, t1, t3), spt(t1, x - 1, t1, t2);
    o = mrg(mrg(t1, mrg(lc[t2], rc[t2])), t3);
}
int rk(int o, int x, int re = 1) {
    while(o) o = x > c[o] ? re += s[lc[o]] + 1, rc[o] : lc[o]; 
    return re;
}
int kth(int o, int k) {
    while(k != s[lc[o]] + 1)
        o = k > s[lc[o]] ? k -= s[lc[o]] + 1, rc[o] : lc[o];
    return c[o];
}
int pre(int o, int x, int re = -inf) {
    while(o) o = x > c[o] ? re = max(re, c[o]), rc[o] : lc[o];
    return re;
}
int nxt(int o, int x, int re = inf) {
    while(o) o = x < c[o] ? re = min(re, c[o]), lc[o] : rc[o];
    return re;
}

可持久化版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
#include <cstdio>
#include <limits>
#include <vector>
using namespace std;

typedef long long int64;

inline int64 Fingerprint(int64 x) {
    const int64 kMul = 0x9ddfea08eb382d69ULL;
    x *= kMul, x ^= x >> 47;
    x *= kMul, x ^= x >> 47;
    x *= kMul, x ^= x >> 47;
    return x * kMul;
}

inline int64 Random() {
    static int64 Seed = 2;
    Seed += Fingerprint(Seed);
    return Seed & numeric_limits<int64>::max();
}

typedef int DataType;

struct RBST {
    RBST *ChildL, *ChildR;
    int Size;
    DataType Data;

    RBST() { ChildL = ChildR = 0; }
};

int GetSize(const RBST* root) { return root ? root->Size : 0; }

int LowerBoundIndex(const RBST* root, DataType x) {
    if (!root) return 0;
    if (x <= root->Data) return LowerBoundIndex(root->ChildL, x);
    int sizeL = GetSize(root->ChildL);
    return LowerBoundIndex(root->ChildR, x) + sizeL + 1;
}

DataType Select(const RBST* root, int index) {
    int sizeL = GetSize(root->ChildL);
    if (index == sizeL) return root->Data;
    if (index < sizeL) return Select(root->ChildL, index);
    return Select(root->ChildR, index - sizeL - 1);
}

RBST* SetSize(RBST* root) {
    root->Size = GetSize(root->ChildL) + GetSize(root->ChildR) + 1;
    return root;
}

struct RBSTree {
    vector<RBST*> Nodes;

    RBST* NewNode(RBST* node) {
        Nodes.push_back(new RBST(*node));
        return Nodes.back();
    }

    void Split(RBST* root, DataType x, RBST*& treeL, RBST*& treeR) {
        if (!root) {
            treeL = treeR = 0;
        } else if (x <= root->Data) {
            RBST *newRoot = NewNode(root);
            Split(root->ChildL, x, treeL, newRoot->ChildL);
            treeR = SetSize(newRoot);
        } else {
            RBST* newRoot = NewNode(root);
            Split(root->ChildR, x, newRoot->ChildR, treeR);
            treeL = SetSize(newRoot);
        }
    }

    RBST* Join(RBST* treeL, RBST* treeR) {
        int sizeL = GetSize(treeL);
        int sizeR = GetSize(treeR);
        int size = sizeL + sizeR;
        if (size == 0) return 0;
        if (Random() % size < sizeL) {
            RBST* newRoot = NewNode(treeL);
            newRoot->ChildR = Join(treeL->ChildR, treeR);
            return SetSize(newRoot);
        } else {
            RBST* newRoot = NewNode(treeR);
            newRoot->ChildL = Join(treeL, treeR->ChildL);
            return SetSize(newRoot);
        }
    }

    RBST* InsertAsRoot(RBST* root, DataType item) {
        Nodes.push_back(new RBST);
        RBST *newRoot = Nodes.back();
        newRoot->Data = item;
        Split(root, item + 1, newRoot->ChildL, newRoot->ChildR);
        return SetSize(newRoot);
    }

    RBST* Insert(RBST* root, DataType item) {
        if (Random() % (GetSize(root) + 1) == 0) {
            return InsertAsRoot(root, item);
        } else if (item < root->Data) {
            RBST *newRoot = NewNode(root);
            newRoot->ChildL = Insert(root->ChildL, item);
            return SetSize(newRoot);
        } else {
            RBST *newRoot = NewNode(root);
            newRoot->ChildR = Insert(root->ChildR, item);
            return SetSize(newRoot);
        }
    }

    RBST* Remove(RBST* root, DataType item) {
        RBST *tree1, *tree2, *tree3, *tree4 = 0;
        Split(root, item, tree1, tree2);
        Split(tree2, item + 1, tree2, tree3);
        if (tree2) tree4 = Join(tree2->ChildL, tree2->ChildR);
        return Join(Join(tree1, tree4), tree3);
    }

    void Destroy() {
        for (int i = 0; i < Nodes.size(); ++i) delete Nodes[i];
    }
};

K-D 树

定义

1
2
3
4
5
6
7
8
9
10
11
12
13
14
#define mid ((l + r) >> 1)
#define lc (o << 1)
#define rc (o << 1 | 1)
#define lch l, mid, lc
#define rch mid + 1, r, rc

int D, L[N * 4][K], R[N * 4][K];
struct node {
    int x[K];
    void read() { For(i, 0, K) scanf("%d", &x[i]); }
    bool operator <(const node& b)const {
        return x[D] < b.x[D];
    }
} a[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
lf sq(lf x) { return x * x; }
void pu(int o) {
    For(i, 0, K) {
        L[o][i] = min(L[lc][i], L[rc][i]);
        R[o][i] = max(R[lc][i], R[rc][i]);
    }
}
void bld(int l, int r, int o) {
    if(l == r) {
        For(i, 0, K) L[o][i] = R[o][i] = a[l].x[i];
        return;
    }
    lf va[K] = {};
    For(i, 0, K) {
        lf av = 0;
        rep(j, l, r) av += a[j].x[i];
        av /= r - l + 1;
        rep(j, l, r) va[i] += sq(a[j].x[i] - av);
    }
    D = max_element(va, va + K) - va;
    nth_element(a + l, a + mid, a + r + 1);
    bld(lch), bld(rch), pu(o);
}

欧几里得距离平方

1
2
3
4
5
6
7
8
9
ll sq(int x) { return 1ll * x * x; }
ll mine(int o, int x[], ll re = 0) {
    For(i, 0, K) re += sq(max(L[o][i] - x[i], 0)) + sq(max(x[i] - R[o][i], 0));
    return re;
}
ll maxe(int o, int x[], ll re = 0) {
    For(i, 0, K) re += max(sq(x[i] - L[o][i]), sq(x[i] - R[o][i]));
    return re;
}

曼哈顿距离

1
2
3
4
5
6
7
8
int minm(int o, int x[], int re = 0) {
    For(i, 0, K) re += max(L[o][i] - x[i], 0) + max(x[i] - R[o][i], 0);
    return re;
}
int maxm(int o, int x[], int re = 0) {
    For(i, 0, K) re += max(abs(x[i] - L[o][i]), abs(x[i] - R[o][i]));
    return re;
}

切比雪夫距离

1
2
3
4
5
6
7
8
int minc(int o, int x[], int re = inf) {
    For(i, 0, K) re = max(re, max(L[o][i] - x[i], 0) + max(x[i] - R[o][i], 0));
    return re;
}
int maxc(int o, int x[], int re = 0) {
    For(i, 0, K) re = max({re, abs(x[i] - L[o][i]), abs(x[i] - R[o][i])});
    return re;
}

最近点查询

定义

1
ll res;

函数

1
2
3
4
5
6
7
8
9
10
11
void qry(int x[], int l, int r, int o) {
    if(l == r) return void(res = mine(o, x));
    ll dl = mine(lc, x), dr = mine(rc, x);
    if(dl < dr) {
        if(dl < res) qry(x, lch);
        if(dr < res) qry(x, rch);
    } else {
        if(dr < res) qry(x, rch);
        if(dl < res) qry(x, lch);
    }
}

初始化

1
res = inf;

最远点查询

定义

1
ll res;

函数

1
2
3
4
5
6
7
8
9
10
11
void qry(int x[], int l, int r, int o) {
    if(l == r) return void(res = maxe(o, x));
    ll dl = maxe(lc, x), dr = maxe(rc, x);
    if(dl > dr) {
        if(dl > res) qry(x, lch);
        if(dr > res) qry(x, rch);
    } else {
        if(dr > res) qry(x, rch);
        if(dl > res) qry(x, lch);
    }
}

初始化

1
res = 0;

矩形判定

矩形是否包含所有点

1
2
3
4
int chk1(int l[], int r[], int o, int re = 1) {
    For(i, 0, K) re &= l[i] <= L[o][i] && R[o][i] <= r[i];
    return re;
}

矩形是否可能包含点

1
2
3
4
int chk2(int l[], int r[], int o, int re = 1) {
    For(i, 0, K) re &= max(l[i], L[o][i]) <= min(r[i], R[o][i]);
    return re;
}

圆判定

圆是否包含所有点

1
2
3
int chk1(int x[], int r, int o) {
    return maxe(x, o) <= 1ll * r * r;
}

圆是否可能包含点

1
2
3
int chk2(int x[], int r, int o) {
    return mine(x, o) <= 1ll * r * r;
}

图论

虚树 / vtree

vector 版

定义

1
2
3
int n, a[N], tp, stk[N];
int idx, dfn[N], d[N], fa[20][N];
vector<int> G[N], T[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
void dfs(int u) {
    dfn[u] = ++idx;
    rep(i, 1, 19) fa[i][u] = fa[i - 1][fa[i - 1][u]];
    for(int v : G[u]) if(v != fa[0][u])
        d[v] = d[u] + 1, fa[0][v] = u, dfs(v);
}
int lca(int u, int v) {
    if(d[u] < d[v]) swap(u, v);
    rep(i, 0, 19) if(d[u] - d[v] >> i & 1) u = fa[i][u];
    if(u == v) return u;
    per(i, 19, 0) if(fa[i][u] != fa[i][v]) u = fa[i][u], v = fa[i][v];
    return fa[0][u];
}
void bld(int k) {
    sort(a + 1, a + k + 1, [](int a, int b) { return dfn[a] < dfn[b]; });
    stk[++tp] = 1;
    rep(i, 1, k) if(a[i] > 1) {
        int x = lca(a[i], stk[tp]);
        while(d[stk[tp]] > d[x]) tp--, T[d[stk[tp]] > d[x] ? stk[tp] : x].pb(stk[tp + 1]);
        if(stk[tp] != x) stk[++tp] = x;
        stk[++tp] = a[i];
    }
    while(--tp) T[stk[tp]].pb(stk[tp + 1]);
}
void clr(int u) {
    for(int v : G[u]) clr(v); T[u].clear();
}

快很多版

定义

1
2
3
4
5
int n, a[N], p, stk[N];
int d[N], fa[N], sz[N], son[N], tp[N], idx, dfn[N];
vector<int> G[N];
int eid, he[N];
struct edge { int v, n; } e[N * 2];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
void dfs(int u) {
    sz[u] = 1;
    for(int v : G[u]) {
        d[v] = d[u] + 1, dfs(v), sz[u] += sz[v];
        if(sz[v] > sz[son[u]]) son[u] = v;
    }
}
void Dfs(int u, int top) {
    tp[u] = top, dfn[u] = ++idx;
    if(son[u]) Dfs(son[u], top);
    for(int v : G[u]) if(v != son[u]) Dfs(v, v);
}
int lca(int u, int v) {
    for(; tp[u] != tp[v]; d[tp[u]] > d[tp[v]] ? u = fa[tp[u]] : v = fa[tp[v]]);
    return d[u] < d[v] ? u : v;
}
void add(int u, int v, int t = 1) {
    e[++eid] = {v, h1[u]}, h1[u] = eid;
}
void bld(int k) {
    sort(a + 1, a + k + 1, [](int a, int b) { return dfn[a] < dfn[b]; });
    stk[++p] = 1;
    rep(i, 1, k) if(a[i] > 1) {
        int x = lca(a[i], stk[p]);
        while(d[stk[p]] > d[x])
            p--, add(d[stk[p]] > d[x] ? stk[p] : x, stk[p + 1]);
        if(stk[p] != x) stk[++p] = x;
        stk[++p] = a[i];
    }
    while(--p) add(stk[p], stk[p + 1]);
}

超快版倍增求 LCA

定义

1
int dl[N], dr[N], fa[20][N];

函数

1
2
3
4
5
6
7
8
9
int lca(int u, int v) {
    if(dl[u] < dl[v]) swap(u, v);
    if(dl[v] <= dl[u] && dl[u] <= dr[v]) return v;
    per(i, 19, 0) {
        int x = fa[i][u];
        if(x && dl[x] > dl[v]) u = x;
    }
    return fa[0][u];
}

最大流 / mf

int 版

定义

1
2
3
int n, m, s, t;
int eid = 1, he[N], nw[N], d[N], q[N];
struct edge { int v, n, c; } e[M * 2];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
void add(int u, int v, int c) {
    e[++eid] = {v, he[u], c}, he[u] = eid;
    e[++eid] = {u, he[v], 0}, he[v] = eid;
}
bool bfs() {
    mem(d, 0), memcpy(nw, he, sizeof he);
    q[1] = s, d[s] = 1;
    for(int l = 1, r = 1; l <= r; l++) {
        int u = q[l];
        for(int i = he[u], v; v = e[i].v; i = e[i].n)
        if(e[i].c && !d[v]) q[++r] = v, d[v] = d[u] + 1;
    }
    return d[t];
}
int dfs(int u, int lim, int re = 0) {
    if(u == t) return lim;
    for(int& i = nw[u], v; v = e[i].v; i = e[i].n)
    if(e[i].c && d[v] == d[u] + 1) {
        int t = dfs(v, min(e[i].c, lim));
        e[i].c -= t, e[i ^ 1].c += t, re += t, lim -= t;
        if(!lim) break;
    }
    return re;
}

使用

1
2
int flow = 0;
while(bfs()) flow += dfs(s, inf);

long long 版

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
bool bfs() {
    mem(d, 0), memcpy(nw, he, sizeof he);
    q[1] = s, d[s] = 1;
    for(int l = 1, r = 1; l <= r; l++) {
        int u = q[l];
        for(int i = he[u], v; v = e[i].v; i = e[i].n)
        if(e[i].c && !d[v]) q[++r] = v, d[v] = d[u] + 1;
    }
    return d[t];
}
ll dfs(int u, ll lim, ll re = 0) {
    if(u == t) return lim;
    for(int& i = nw[u], v; v = e[i].v; i = e[i].n)
    if(e[i].c && d[v] == d[u] + 1) {
        ll t = dfs(v, min((ll)e[i].c, lim));
        e[i].c -= t, e[i ^ 1].c += t, re += t, lim -= t;
        if(!lim) break;
    }
    return re;
}

最小费用最大流 / mcf

SSP 算法

定义

1
2
3
int n, m, s, t, flow, cost;
int eid = 1, he[N], nw[N], q[N], d[N], vis[N];
struct edge { int v, n, c, w; } e[M * 2];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
void add(int u, int v, int c, int w) {
    e[++eid] = {v, he[u], c, w}, he[u] = eid;
    e[++eid] = {u, he[v], 0, -w}, he[v] = eid;
}
bool spfa() {
    mem(d, 63), memcpy(nw, he, sizeof he);
    q[0] = s, d[s] = 0, vis[s] = 1;
    for(int l = 0, r = 0; l <= r; l++) {
        int u = q[l % N]; vis[u] = 0;
        for(int i = he[u], v; v = e[i].v; i = e[i].n)
        if(e[i].c && d[u] + e[i].w < d[v]) {
            d[v] = d[u] + e[i].w;
            if(!vis[v]) q[++r % N] = v, vis[v] = 1;
        }
    }
    return d[t] < d[0];
}
int dfs(int u, int lim, int re = 0) {
    if(u == t) return lim;
    vis[u] = 1;
    for(int& i = nw[u], v; v = e[i].v; i = e[i].n)
    if(!vis[v] && e[i].c && d[v] == d[u] + e[i].w) {
        int t = dfs(v, min(lim, e[i].c));
        e[i].c -= t, e[i ^ 1].c += t, re += t, lim -= t, cost += t * e[i].w;
        if(!lim) break;
    }
    vis[u] = 0;
    return re;
}

使用

1
while(spfa()) flow += dfs(s, 1e9);

Primal-Dual 原始对偶算法

定义

1
2
3
4
5
6
7
8
9
int n, m, s, t, flow, cost;
int eid = 1, he[N], d[N], pre[N], h[N];
struct edge { int v, n, c, w; } e[M * 2];
struct node {
    int d, u;
    bool operator <(const node& b)const {
        return d > b.d;
    }
};

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
void add(int u, int v, int c, int w) {
    e[++eid] = {v, he[u], c, w}, he[u] = eid;
    e[++eid] = {u, he[v], 0, -w}, he[v] = eid;
}
bool dij() {
    priority_queue<node> q;
    mem(d, 63), q.push({d[s] = 0, s});
    while(q.size()) {
        auto [dis, u] = q.top(); q.pop();
        if(dis > d[u]) continue;
        for(int i = he[u], v; v = e[i].v; i = e[i].n) if(e[i].c) {
            int w = d[u] + e[i].w - h[v] + h[u];
            if(w < d[v]) pre[v] = i ^ 1, q.push({d[v] = w, v});
        }
    }
    rep(i, 1, /* Number of nodes */) h[i] += d[i];
    return d[t] < d[0];
}

使用

1
2
3
4
5
6
while(dij()) {
    int mn = d[0];
    for(int u = t, i; i = pre[u]; u = e[i].v) mn = min(mn, e[i ^ 1].c);
    for(int u = t, i; i = pre[u]; u = e[i].v) e[i].c += mn, e[i ^ 1].c -= mn;
    flow += mn, cost += mn * h[t];
}

二分图最大匹配 / match

定义

1
2
int n, m, vis[N], mat[N];
vector<int> G[N];

函数

1
2
3
4
5
6
int dfs(int u, int s) {
    if(vis[u] == s) return 0;
    vis[u] = s;
    for(int v : G[u]) if(!mat[v] || dfs(mat[v], s)) return mat[v] = u;
    return 0;
}

使用

1
2
int as = 0;
rep(i, 1, n) if(dfs(i, i)) as++;

2-SAT 问题

已弃用。（复杂度不对）

定义

1
2
int n, m, co[N * 2], stk[N * 2], tp;
vector<int> G[N * 2];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
void add(int u, int a, int v, int b) {
    G[u * 2 + !a].pb(v * 2 + b);
    G[v * 2 + !b].pb(u * 2 + a);
}
int dfs(int u) {
    if(co[u] | co[u ^ 1]) return co[u];
    co[u] = 1, stk[++tp] = u;
    for(int v : G[u]) if(!dfs(v)) return 0;
    return 1;
}
int twoSat() {
    rep(i, 1, n) {
        if(!co[i * 2] && !co[i * 2 + 1] && !dfs(i * 2)) {
            while(tp) co[stk[tp--]] = 0;
            if(!dfs(i * 2 + 1)) return 0;
        }
        tp = 0;
    }
    return 1;
}

字符串

manacher 求偶回文串

1
2
3
4
5
rep(i, 1, n) {
    int& j = R[i] = ma > i ? min(R[p * 2 - i], ma - i) : 0;
    while(s[i - j] == s[i + j + 1]) j++;
    if(i + j > ma) ma = i + j, p = i;
}

回文自动机 PAM

普通版

定义

1
2
char s[N];
int n, sz = 1, nw, len[N], f[N], ch[N][26];

函数

1
2
3
4
5
6
7
8
9
10
11
12
void ins(int i) {
    auto jmp = [&](int o) {
        while(s[i - len[o] - 1] != s[i]) o = f[o];
        return o;
    };
    int o = jmp(nw), c = s[i] - 97;
    if(!ch[o][c]) {
        f[++sz] = ch[jmp(f[o])][c];
        len[ch[o][c] = sz] = len[o] + 2;
    }
    nw = ch[o][c];
}

预处理

1
len[1] = -1, f[0] = 1;

偶回文版

函数

1
2
3
4
5
6
7
8
9
10
11
12
void ins(int i) {
    auto jmp = [&](int o) {
        while(o && s[i - len[o] - 1] != s[i]) o = f[o];
        return o;
    };
    int o = jmp(nw), c = s[i] - 97;
    if(!ch[o][c]) {
        f[++sz] = ch[jmp(f[o])][c];
        len[ch[o][c] = sz] = len[o] + 2;
    }
    nw = ch[o][c];
}

预处理

1
rep(i, 0, 25) ch[0][i] = 1;

后缀数组 / SA

定义

1
2
3
int n, sa[N], rk[N], tp[N], px[N], buc[N];
int h[20][N];
char s[N];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
void SA() {
    int m = 128;
    rep(i, 1, n) buc[rk[i] = s[i]]++;
    rep(i, 1, m) buc[i] += buc[i - 1];
    per(i, n, 1) sa[buc[rk[i]]--] = i;
    for(int k = 1, p; memset(buc, p = 0, m + 1 << 2); k <<= 1) {
        rep(i, n - k + 1, n) p++, px[p] = rk[tp[p] = i];
        rep(i, 1, n) if(sa[i] > k) p++, px[p] = rk[tp[p] = sa[i] - k];
        rep(i, 1, n) buc[rk[i]]++;
        rep(i, 1, m) buc[i] += buc[i - 1];
        per(i, n, 1) sa[buc[px[i]]--] = tp[i];
        memcpy(tp, rk, n + 1 << 2), p = 0;
        rep(i, 1, n) {
            int a = sa[i], b = sa[i - 1];
            rk[a] = p += tp[a] != tp[b] || tp[a + k] != tp[b + k];
        }
        if((m = p) >= n) break;
    }
}
void height() {
    int p = 0;
    rep(i, 1, n) {
        for(p && p--; s[i + p] == s[sa[rk[i] - 1] + p]; p++);
        h[0][rk[i]] = p;
    }
    rep(i, 1, 19) rep(j, 1 << i, n)
        h[i][j] = min(h[i - 1][j], h[i - 1][j - (1 << i - 1)]);
}

后缀自动机 / SAM

定义

1
2
char s[N];
int n, sz = 1, nw = 1, f[N * 2], len[N * 2], ch[N * 2][26];

函数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
void ins(int c) {
    int u = ++sz;
    len[u] = len[nw] + 1;
    while(nw && !ch[nw][c]) ch[nw][c] = u, nw = f[nw];
    if(!nw) f[u] = 1;
    else {
        int v = ch[nw][c];
        if(len[v] > len[nw] + 1) {
            f[++sz] = f[v], memcpy(ch[sz], ch[v], sizeof ch[v]);
            f[v] = f[u] = sz, len[sz] = len[nw] + 1;
            while(ch[nw][c] == v) ch[nw][c] = sz, nw = f[nw];
        } else f[u] = v;
    }
    nw = u;
}

最小表示法 / smallest cyclic shift

函数

1
2
3
4
5
6
7
8
9
10
11
int calc(char s[]) {
    int i = 0, j = 1, k = 0;
    while(max({i, j, k}) < n)
    if(s[(i + k) % n] == s[(j + k) % n]) k++;
    else {
        if(s[(i + k) % n] > s[(j + k) % n]) swap(i, j);
        j += k + 1, k = 0;
        if(i == j) j++;
    }
    return min(i, j);
}

最小后缀 / smallest suffix

函数

1
2
3
4
5
6
7
8
9
10
11
int calc(char s[]) {
    int i = 0, j = 1;
    while(s[j]) {
        int k = 0;
        while(s[i + k] == s[j + k]) k++;
        if(!s[j + k]) i = max(i + k, j + 1), swap(i, j);
        else if(s[i + k] > s[j + k]) i = max(i + k, j) + 1, swap(i, j);
        else j += k + 1;
    }
    return i;
}