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
 

 

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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;
}

 

posted @ 2019-07-29 17:32  black_hole6  阅读(388)  评论(0编辑  收藏  举报