小技巧函数

字符数组快速求最小字典

 原理不多说，用笔在纸上模拟一下就会懂的。

int Min_index(char *ss){
    int i = 0,j = 1, k = 0;
    int len = strlen(ss);
    while(i<len && j<len && k<len){
        int t = ss[(j+k)%len]-ss[(i+k)%len];
        if(t == 0)k++;
        else{
            if(t>0) j+=k+1;
            else i+=k+1;
            k=0;
        }
        if(i==j)j++;
    }
    return min(i,j);
}

同理，也可以快速求最大字典

求质因子，在欧拉函数和莫比乌斯函数用的上。

int fun(int x){
    int res = 0;
    for(int i = 2; i <= sqrt(x); i ++){
        if(x%i==0){
            res++;
            while(x%i==0)x/=i;
        }
    }
    if(x > 1) res++;
    return res;
}

乘法逆元。a/b%n == a*k%n == a*inv(b,n);

void exgcd(ll a, ll b, ll &x, ll &y) {
    if(!b){
        x = 1; y = 0;
    }else{
        exgcd(b, a%b, y, x);
        y -= x*(a/b);
    }
}
ll inv(ll a, ll n) {
    ll x, y;
    exgcd(a, n, x, y);
    return (x+n)%n;
}

a*x+b*y=d

void exgcd(ll a, ll b,ll &d, ll &x, ll &y) {
    if(!b){
        d = a;
        x = 1; y = 0;
    }else{
        exgcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}

上面的乘法有误差6*inv(2,mod)不等于3   费马小定理没有

费马小定理求逆元a/b%mod = a*pow_mod(b,mod-2)%mod

ll pow_mod(ll x, ll n){  
    ll res=1;  
    while(n>0){  
        if(n&1)res=res*x%mod;  
        x=x*x%mod;  
        n>>=1;  
    }  
    return res;  
}

 基姆拉尔森公式，计算某日是星期几

int day(int y, int m, int d) {
    if(m == 1 || m == 2) {
        m += 12;
        y --;
    }
    int w = (d + 2*m + 3*(m+1)/5 + y + y/4 - y/100 + y/400) % 7;
    return w;
}

求C(n,m)%p，都是在[1,1e18]范围内

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 15;
ll qmul(ll a,ll p,ll m){
    ll tmp = 0;
    while(p){
        if(1&p) tmp = (tmp+a)%m;
        a = (a+a)%m;
        p>>=1;
    }
    return tmp;
}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){
    if(!b) d=a,x = 1,y=0;
    else exgcd(b,a%b,y,x,d),y-=(a/b)*x;
}
ll inv(ll a,ll p){
    ll x,y,d;
    exgcd(a,p,x,y,d);
    return d==1?(x+p)%p:-1;
}
ll fat(ll x,ll p){
    ll ans = 1;
    for( int i = 2 ; i <= x ; i ++ ) ans = ( ans * i ) % p;
    return ans;
}
ll c(ll n,ll m,ll p){
    if (m < 0 || m > n) return 0; 
    return fat(n,p)*inv(fat(m,p),p)%p*inv(fat(n-m,p),p)%p;
}
ll crt(ll *a,ll *m,int n){
    ll M = 1,res = 0;
    for( int i = 0 ; i < n ; i++ ) M*=m[i];
    for( int i = 0 ; i < n ; i++ ){
        ll w = M/m[i];
        res = (res+qmul(w*inv(w,m[i]),a[i],M))%M;
    }
    return (res+M)%M;
}
ll Lucas(ll n,ll m,ll p){
    return m?Lucas(n/p,m/p,p)*c(n%p,m%p,p)%p:1;
}
int main() {
    ll a[N], p[N];
    ll t, n, m, k;
    cin >> t;
    while(t--) {
        cin >> n >> m >> k;
        for(int i = 0; i < k; i ++) {
            cin >> p[i];
            a[i] = Lucas(n, m, p[i]);
        }
        printf("%lld
",crt(a, p, k));
    }
    return 0;
}