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("*********
")
#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)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"
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("
");
}
}
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
", 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
", 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=0quad and quad C_{n+1}=frac{2(2n+1)}{n+2}C_n\
C_0=1quad 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 ("
The winner is %d
", 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,"
"[!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
", 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:nquad xorquad(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
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