【51NOD-0】1008 N的阶乘 mod P

【算法】简单数学

【题解】多项式展开：(a*b)%p=(a%p*b%p)%p

#include<cstdio>
#include<algorithm>
#define rep(i,j,k) for(int i=j;i<=k;i++)
using namespace std;
int n,p;
int main()
{
    scanf("%d%d",&n,&p);
    int ans=1;
    rep(i,1,n)ans=(1ll*ans*i)%p;
    printf("%d",ans);
    return 0;
}

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转载于:https://www.cnblogs.com/onioncyc/p/6900586.html