排列组合+逆元模板

 

运用费马小定理

#include<stack>
#include<queue>
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fi first
#define se second
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int mod=1e9+7;
int v[maxn];
ll ppow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
ll solve(ll now,ll num)
{
    ll ans=1;
    for(ll i=now;i>now-num;--i)
        ans=(ans*i)%mod;
    for(ll i=1;i<=num;++i)
    {
        ans=(ans*ppow(i,mod-2))%mod;
    }
    return ans;
}
int main()
{
　　printf("%d\n",(ppow(5,mod-2)%mod*4)%mod); //求4/5的逆元
    printf("%d\n",solve(9,3));
        return 0;
}    

 

 

这一个是常数时间求逆元，参考链接：https://blog.csdn.net/qq_36979930/article/details/81698003

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod=1e9+7;
const int maxn=1e5+7;
 
ll fac[maxn];
ll inv[maxn];
ll C(int m,int n)
{
    if(m>n)
        return -1;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
 
ll quick_mod(ll a,ll m)
{
    ll tmp=a%mod;
    ll ans=1;
    while(m)
    {
        if(m&1)
            ans=ans*tmp%mod;
        tmp=tmp*tmp%mod;
        m>>=1;
    }
    return ans;
}
 
void init()
{
    fac[0]=1;
    for(int i=1; i<maxn; i++)
        fac[i]=(fac[i-1]*i)%mod;
    inv[maxn-1]=quick_mod(fac[maxn-1],mod-2);
    for(int i=maxn-2; i>=0; i--)
        inv[i]=(inv[i+1]*(i+1))%mod;
}
 
int main()
{
    init();
    ll n,m;
    while(cin>>n>>m)
        cout << C(m,n) << endl;
    return 0;
}