HDU-6608-Fansblog(威尔逊定理+快速乘)（多校）

Problem Description

Farmer John keeps a website called ‘FansBlog’ .Everyday , there are many people visited this blog.One day, he find the visits has reached P , which is a prime number.He thinks it is a interesting fact.And he remembers that the visits had reached another prime number.He try to find out the largest prime number Q ( Q < P ) ,and get the answer of Q! Module P.But he is too busy to find out the answer. So he ask you for help. ( Q! is the product of all positive integers less than or equal to n: n! = n * (n-1) * (n-2) * (n-3) *… * 3 * 2 * 1 . For example, 4! = 4 * 3 * 2 * 1 = 24 )

Input

First line contains an number T(1<=T<=10) indicating the number of testcases.
Then T line follows, each contains a positive prime number P (1e9≤p≤1e14)

Output

For each testcase, output an integer representing the factorial of Q modulo P.

Sample Input

1 1000000007

Sample Output

328400734

Source

2019 Multi-University Training Contest 3

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代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>typedef long long ll;
using namespace std;
int prime[10000005];
bool vis[10000005];
int cnt =0;
void erla() {memset(vis,false,sizeof(vis));memset(prime,0,sizeof(prime));for(int t=2; t<=10000003; t++) {if(!vis[t]) {prime[cnt++]=t;}for(int j=0; j<cnt&&t*prime[j]<=10000003; j++) {vis[t*prime[j]]=true;if(t%prime[j]==0) {break;}}}
}
inline ll ksc(ll x,ll y,ll mod)
{return (x*y-(ll)((long double)x/mod*y)*mod+mod)%mod;
}
ll ksm(ll x,ll y,ll mod)
{ll ans=1;while(y){if(y&1){ans=ksc(ans,x,mod);}x=ksc(x,x,mod);y>>=1;}return ans;
}
bool isprime(ll x)
{for(int t=0;t<cnt&&prime[t]<x;t++){if(x%prime[t]==0){return false;}}return true;
}
int main()
{int T;cin>>T;erla();while(T--){ll n;scanf("%lld",&n);ll ans=1;for(ll t=n-2;t>=2;t--){if(isprime(t)){break;}ans=ksc(ans,ksm(t,n-2,n),n);//cout<<ans<<endl;
        }printf("%lld\n",ans);}    return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/11265302.html

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