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#include<stdio.h>
using namespace std;
typedef long long ll;
int main(){
ll n,p,ans = 1;
scanf("%lld%lld",&n,&p);
for(int i = 1 ; i <= n ; ++i){
ans = (ans * i) % p;
}
printf("%d\n",ans);
return 0;
}
[2016-05-09][51nod][1008 N的阶乘 mod P]
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原文地址:http://www.cnblogs.com/qhy285571052/p/a05ea631a23371459f59618cbbe68aef.html