buerdepepeqi的模板

头文件

#include <set>

#include <map>

#include <deque>

#include <queue>

#include <stack>

#include <cmath>

#include <ctime>

#include <bitset>

#include <cstdio>

#include <string>

#include <vector>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long LL;

typedef pair<LL, LL> pLL;

typedef pair<LL, int> pLi;

typedef pair<int, LL> pil;;

typedef pair<int, int> pii;

typedef unsigned long long uLL;

#define ls rt<<1

#define rs rt<<1|1

#define lson l,mid,rt<<1

#define rson mid+1,r,rt<<1|1

#define bug printf("*********\n")

#define FIN freopen("input.txt","r",stdin);

#define FON freopen("output.txt","w+",stdout);

#define IO ios::sync_with_stdio(false),cin.tie(0)

#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"

#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"

#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"

LL read() {

    int x = 0, f = 1; char ch = getchar();

    while(ch < '0' || ch > '9') {

        if(ch == '-')f = -1;

        ch = getchar();

    }

    while(ch >= '0' && ch <= '9') {

        x = x * 10 + ch - '0';

        ch = getchar();

    }

    return x * f;

}

const double eps = 1e-8;

const int mod = 1e9+7;

const int maxn = 3e5 + 5;

const int INF = 0x3f3f3f3f;

const LL INFLL = 0x3f3f3f3f3f3f3f3f;

数学

费马小定理

a是不能被质数p整除的正整数，则有a^(p-1)≡ 1 mod p,即a^(p-1) mod p=1;

推论：对于不能被质数p整除的正整数a，有a^p≡ a mod p

a^b%p=a^(b%(p-1))%p

逆元

/*O(n)打表求1~n的逆元*/

LL inv[maxn];

void init() {

    inv[1] = 1;

    for (int i = 2; i < maxn; i++) inv[i] = inv[mod % i] * (mod - mod / i) % mod;

}

矩阵快速幂

struct matrix {//矩阵

    int n;//长

    int m;//宽

    long long a[105][105];

    matrix() {//构造函数

        n = 2;

        m = 2;

        memset(a, 0, sizeof(a));

    }

    matrix(int x, int y) {

        n = x;

        m = y;

        memset(a, 0, sizeof(a));

    }

    void print() {

        for(int i = 1; i <= n; i++) {

            for(int j = 1; j <= m; j++) {

                printf("%d ", a[i][j]);

            }

            printf("\n");

        }

    }

    void setv(int x) {//初始化

        if(x == 0) {

            memset(a, 0, sizeof(a));

        }

        if(x == 1) {

            memset(a, 0, sizeof(a));

            for(int i = 1; i <= n; i++) a[i][i] = 1;

        }

    }

    friend matrix operator *(matrix x, matrix y) { //矩阵乘法

        matrix tmp = matrix(x.n, y.m);

        for(int i = 1; i <= x.n; i++) {

            for(int j = 1; j <= y.m; j++) {

                tmp.a[i][j] = 0;

                for(int k = 1; k <= y.n; k++) {

                    tmp.a[i][j] += (x.a[i][k] * y.a[k][j]) % mod;

                }

                tmp.a[i][j] %= mod;

            }

        }

        return tmp;

    }

};

matrix fast_pow(matrix x, long long k) { //矩阵快速幂

    matrix ans = matrix(n, n);

    ans.setv(1);//初始化为1

    while(k > 0) { //类似整数快速幂

        if(k & 1) {

            ans = ans * x;

        }

        k >>= 1;

        x = x * x;

    }

    return ans;

}

欧拉定理和欧拉函数

欧拉函数phi(m)：当m>1是，phi(m)表示比m小且与m互质的正整数个数

欧拉定理：a^phi(n) = 1 mod n

a^x = a^(x % phi(p) + p) (mod p)

//打表

int b[N],prime[N],phi[N];

void init(){  //求欧拉函数和素数

    int i,j,k,c=0;

    memset(b,0,sizeof(b));

    for(i=2;i<N;i++){

        if(b[i]==0){

            prime[++c]=i;

            phi[i]=i-1;

        }

        for(j=1;(j<=c)&&(k=i*prime[j])<N;j++){

            b[k]=1;

            if(i%prime[j])  phi[prime[j]*i]=phi[i]*(prime[j]-1);

            else phi[prime[j]*i]=phi[i]*prime[j];

        }

    }

}

//直接计算欧拉函数

int phi (int n) {

    int res = n;

    for (int i = 2; i * i <= n; ++i) {

        if (n % i == 0) {

            res -= res / i;

            while (n % i == 0) n /= i;

        }

    }

    return n > 1 ? res / n * (n - 1) : res;

}

//欧拉函数递归调用

//计算dp[i] = 2^dp[i+1] % mod，需要先计算mod的phi，以及phi(mod)的phi直到为1

//dp[i] = 2^(dp[i+1]%phi(mod)+phi(mod)) %mod

const int MX = 1e5 + 5;

const ll mod = 1e9 + 7;

ll m[50], dp[MX][50];

int sz, n;

int Phi (int n) {

    int res = n;

    for (int i = 2; i * i <= n; ++i) {

        if (n % i == 0) {

            res -= res / i;

            while (n % i == 0) n /= i;

        }

    }

    return n > 1 ? res / n * (n - 1) : res;

}

void init() {

    int x = mod;

    sz = 0;

    m[++sz] = x;

    while (x != 1) m[++sz] = Phi (x), x = m[sz], cout << x << endl;

}

ll pow (ll a, ll b, ll m) {

    ll ret = 1;

    while (b) {

        if (b & 1) ret = ret * a % m;

        a = a * a % m;

        b >>= 1;

    }

    return ret;

}

int main() {

    init();

    for (int i = 1; i <= n; i++) {

        for (int j = 1; j < sz; j++) {

            dp[i][j] = pow (2ll, dp[i - 1][j + 1], m[j]) * 3 - 3;

            while (dp[i][j] < 0) dp[i][j] += m[j];

            while (dp[i][j] >= m[j]) dp[i][j] -= m[j];

        }

    }

    printf ("%lld\n", dp[n][1] % m[1]);

}

扩展欧几里得定理

//递归

LL ex_gcd(LL a, LL b, LL &x, LL &y) {

    if(b == 0) {

        x = 1, y = 0;

        return a;

    }

    int d = ex_gcd(b, a % b, x, y);

    int z = x;

    x = y;

    y = z - y * (a / b);

    return d;

}

//非递归

//返回值为 gcd 以及 a,b 的系数

pair<ll, PLL> exgcd(ll a,ll b){

    valarray<ll>equation[2] = {{a,1,0},{b,0,1}};

    while (equation[1][0]) {

        ll q = equation[0][0]/equation[1][0];

        equation[0] -= q * equation[1];

        swap(equation[0], equation[1]);

    }

    return MP(equation[0][0],MP(equation[0][1],equation[0][2]));

}

组合数

1、普通组合数

void gcd(LL a, LL b, LL &d, LL &x, LL &y) {

    if(!b) {

        d = a;

        x = 1;

        y = 0;

    } else {

        gcd(b, a % b, d, y, x);

        y -= x * (a / b);

    }

}

LL inv(LL a, LL n) {

    LL d, x, y;

    gcd(a, n, d, x, y);

    return (x % n + n) % n;

}

LL fac[maxn], fav[maxn];

LL C(LL n, LL m) {

    return fac[n] * fav[m] % mod * fav[n - m] % mod;

}

void init() {

    fac[0] = 1; fav[0] = 1;

    for(int i = 1; i < maxn; i++)fac[i] = fac[i - 1] * i % mod;

    for(int i = 1; i < maxn; i++)fav[i] = inv(fac[i], mod);

}

2、Lucas组合数

LL p;

LL n, m;

LL fac[maxn];

LL pow(LL y, int z, int p) {

    y %= p; LL ans = 1;

    for(int i = z; i; i >>= 1, y = y * y % p)if(i & 1)ans = ans * y % p;

    return ans;

}

LL C(LL n, LL m) {

    if(m > n)return 0;

    return ((fac[n] * pow(fac[m], p - 2, p)) % p * pow(fac[n - m], p - 2, p) % p);

}

LL Lucas(LL n, LL m) {

    if(!m)return 1;

    return C(n % p, m % p) * Lucas(n / p, m / p) % p;

}

int main() {

    scanf("%lld%lld%lld", &n, &m, &p);

    fac[0] = 1;

    for(int i = 1; i <= p; i++) {

        fac[i] = fac[i - 1] * i % p;

    }

    printf("%lld\n", Lucas(n, m) );

    return 0;

}

3、卡特兰数

卡特兰数是一种经典的组合数，经常出现在各种计算中，其前几项为 :

1, 2, 5, 14, 42,

132, 429, 1430, 4862, 16796,

58786, 208012, 742900, 2674440, 9694845,

35357670, 129644790, 477638700, 1767263190,

6564120420, 24466267020, 91482563640, 343059613650, 1289904147324,

4861946401452, ...

\[C_n=\frac{(2n)!}{(n+1)!n!}
\]

卡特兰数满足如下递归

\[C_0=0\quad and \quad C_{n+1}=\frac{2(2n+1)}{n+2}C_n\\
C_0=1\quad and \quad C_{n+1}=\sum_{i=0}^{n}C_iC_{n-i}\quad n>=0
\]

卡特兰数的渐进增长为

\[C_n=> \frac{4^n}{n^{3/2}\sqrt{\pi}}
\]

所有的奇卡塔兰数Cn都满足n = 2^k − 1。

所有其他的卡塔兰数都是偶数。

卡特兰数的经典问题

n个左括号与n个右括号组成的合法序列的数量为卡特兰数
1，2，···，n经过一个栈，形成的合法的出栈序列的数量为卡特兰数
n个节点构成不同的二叉树的数量为卡特兰数
在平面直角坐标系上，每一步只能向上走或向右走，从（0，0）走到（n，n）并且除了两个端点外不能接触直线y=x的路线数量为2*Cat_n-1
Cn表示通过连结顶点而将n + 2边的凸多边形分成三角形的方法个数
Cn表示集合{1, …, n}的不交叉划分的个数. 其中每个段落的长度为2
Cn表示用n个长方形填充一个高度为n的阶梯状图形的方法个数

下图为 n = 4的情况:

４、错排公式

\[D(n)=(n-1)*[D(n-1) +D(n-2)]
\]

约瑟夫环

//最后一个报号的人

int main(){

    int n, m, i, s = 0;

    scanf("%d%d", &n, &m);

    for (i = 2; i <= n; i++)

        s = (s + m) % i;

    printf ("\nThe winner is %d\n", s+1);

}

//第k个出去的人

#include<bits/stdc++.h>

using namespace std;

int main(){

    int n,k;

    while(scanf("%d%d",&n,&k)==2){//n个人第k个人走，最后一个人是谁

        int t; scanf("%d",&t);

        while(t--){

            int q; scanf("%d",&q);

            long long N = (long long)q*k;

            while(N>n){

                N = (N-n-1)/(k-1)+N-n;

            }

            printf("%d%c",(int)N," \n"[!t]);

        }

    }

    return 0;

}

莫比乌斯函数

//线性筛求莫比乌斯函数

int mu[maxn];

int prime[maxn];

int not_prime[maxn];

int n,tot;

void Mobiwus(){

    mu[1] = 1;

    for(int i = 2; i <= n; i++) {

        if(!not_prime[i]) {

            prime[++tot] = i;

            mu[i] = -1;

        }

        for(int j = 1; prime[j]*i <= n; j++) {

            not_prime[prime[j]*i] = 1;

            if(i % prime[j] == 0) {

                mu[prime[j]*i] = 0;

                break;

            }

            mu[prime[j]*i] = -mu[i];

        }

    }

}

BM递推式

namespace linear_seq {

#define pb push_back

#define SZ(x) ((int)(x).size())

#define rep(i,a,b) for(int i=(a);i<(b);++i)

#define per(i,a,b) for(int i=(a)-1;i>=(b);--i)

    typedef vector<int> VI;

    const int N = 10010;

    const LL mod = 1e9 + 7;

    LL res[N], base[N], cnt[N], val[N];

    LL power(LL a, LL b) {

        LL ret = 1; a %= mod;

        while(b) {

            if(b & 1)ret = ret * a % mod;

            a = a * a % mod;

            b >>= 1;

        }

        return ret;

    }

    vector<int> v;

    void mul(LL *a, LL *b, int k) {

        rep(i, 0, k + k) cnt[i] = 0;

        rep(i, 0, k) if (a[i]) rep(j, 0, k) cnt[i + j] = (cnt[i + j] + a[i] * b[j]) % mod;

        per(i, k + k, k) if (cnt[i]) rep(j, 0, SZ(v))

            cnt[i - k + v[j]] = (cnt[i - k + v[j]] - cnt[i] * val[v[j]]) % mod;

        rep(i, 0, k) a[i] = cnt[i];

    }

    int solve(LL n, VI a, VI b) { // a系数 b初值 b[n+1]=a[0]*b[n]+...

        LL ans = 0, pnt = 0;

        int k = SZ(a);

        rep(i, 0, k) val[k - 1 - i] = -a[i]; val[k] = 1;

        v.clear();

        rep(i, 0, k) if (val[i] != 0) v.push_back(i);

        rep(i, 0, k) res[i] = base[i] = 0;

        res[0] = 1;

        while ((1LL << pnt) <= n) pnt++;

        per(p, pnt + 1, 0) {

            mul(res, res, k);

            if ((n >> p) & 1) {

                per(i, k, 0) res[i + 1] = res[i]; res[0] = 0;

                rep(j, 0, SZ(v)) res[v[j]] = (res[v[j]] - res[k] * val[v[j]]) % mod;

            }

        }

        rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;

        if (ans < 0) ans += mod;

        return ans;

    }

    VI BM(VI s) {

        VI C(1, 1), B(1, 1);

        int L = 0, m = 1, b = 1;

        rep(n, 0, SZ(s)) {

            LL d = 0;

            rep(i, 0, L + 1) d = (d + (LL)C[i] * s[n - i]) % mod;

            if (d == 0) ++m;

            else if (2 * L <= n) {

                VI T = C;

                LL c = mod - d * power(b, mod - 2) % mod;

                while (SZ(C) < SZ(B) + m) C.pb(0);

                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;

                L = n + 1 - L; B = T; b = d; m = 1;

            } else {

                LL c = mod - d * power(b, mod - 2) % mod;

                while (SZ(C) < SZ(B) + m) C.pb(0);

                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;

                ++m;

            }

        }

        return C;

    }

    int gao(VI a, LL n) {

        VI c = BM(a);

        c.erase(c.begin());

        rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;

        return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));

    }

};

//用法vector<int>v;

//打表填系数递推

//linear_seq::gao(v, n - 1)

字符串

马拉车

struct Manacher {

    int p[maxn << 1], str[maxn << 1];

    bool check(int x) {

        return x - (x & -x) == 0;

    }

    bool equal(int x, int y) {

        if ((x & 1) != (y & 1)) return false;

        return (x & 1) || (check(mask[x >> 1]) && check(mask[y >> 1]) && str[x] == str[y]);

    }

    int solve(int len) {

        int max_right = 0, pos = 0, ret = 0;

        for (int i = 1; i <= len; i++) {

            p[i] = (max_right > i ? std::min(p[pos + pos - i], max_right - i) : 1);

            if ((i & 1) || check(mask[i >> 1])) {

                while (equal(i + p[i], i - p[i])) p[i]++;

                if (max_right < i + p[i]) pos = i, max_right = i + p[i];

                ret += p[i] / 2;

            } else p[i] = 2;

        }

        return ret;

    }

} manacher;

图论

三元环

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

vector<int> g[N];

int deg[N], vis[N], n, m, ans;

struct E { int u, v; } e[N * 3];

int main() {

    scanf("%d%d", &n, &m);

    for(int i = 1 ; i <= m ; ++ i) {

        scanf("%d%d", &e[i].u, &e[i].v);

        ++ deg[e[i].u], ++ deg[e[i].v];

    }

    for(int i = 1 ; i <= m ; ++ i) {

        int u = e[i].u, v = e[i].v;

        if(deg[u] < deg[v] || (deg[u] == deg[v] && u > v)) swap(u, v);

        g[u].push_back(v);

    }

    for(int x = 1 ; x <= n ; ++ x) {

        for(auto y: g[x]) vis[y] = x;

        for(auto y: g[x])

            for(auto z: g[y])

                if(vis[z] == x)

                    ++ ans;

    }

    printf("%d\n", ans);

}

小技巧

二进制状态压缩

\[取出整数n在二进制表示下的第k位：(n>>k)\&1\\
取出整数n在二进制表示下的第0 \sim k-1位（后k位）:n\&((1<<k)-1)\\
把整数n在二进制表示下第k位取反:n\quad xor\quad(1<<k)\\
对整数n在二进制下表示的第k位赋值1：n|(1<<k)\\
对整数n在二进制表示下的第k为赋值0：n\&(\sim(1<<k))\\
\]

rope的用法

头文件：#include <ext/rope>

命名空间using namespace __gnu_cxx;

定义:rope r;

insert(pos, s)将字符串s插入pos位置

erase(pos, num)从pos开始删除num个字符

copy(pos, len, s)从pos开始len个字符用s代替

substr(pos, len)提取pos开始的len个字符

at(x)访问第x个元素

巴特西

buerdepepeqi 的模版