Lucas定理模版、

资料：传送门

相关题目：传送门

求C（n，m）模p的结果

#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;ll n,m,p;ll quick_mod(ll a, ll b){ll ans = 1;a=a%p;while(b){if(b&1)ans = ans*a%p;b>>=1;a=a*a%p;}return ans;}ll C(ll n, ll m){if(m>n)return 0;if(n==m)return 1;if(m>n-m)m = n-m;ll ans=1, cns=1;for(int i=1; i<=m; ++i){ans=(ans*(n+i-m))%p;cns=(cns*i)%p;}ll res = (ans*quick_mod(cns, p-2))%p;return res;}ll Lucas(ll n, ll m){if(m==0)return 1;return C(n%p, m%p)%p * Lucas(n/p, m/p)%p;}int main(){scanf("%lld%lld%lld",&n,&m,&p);printf("%lld\n", Lucas(n, m));return 0; }