Fansblog （HDU - 6608）（威尔迅定理+费马小定理）

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

题意：T组样例，每组样例，给出一个素数P(1e9≤P≤1e14)，Q是P的前一个素数求Q！%P。

思路：在初等数论中，威尔迅定理给出了判定一个自然数是否为素数的充分必要条件。即：当且仅当p为素数的时候：( p -1 )!  -1 ≡( mod p )，但是由于阶乘是呈爆炸增长的，其结论对于实际操作意义不大。

所以：由威尔迅定理得：((p−1)!)%p​=q!×(q+1)×(q+2)×...×(p−1)%p=p−1​

                                         q!​≡(q+1)×(q+2)×...×(p−2)1​(mod p)​

AC代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e7+10;
ll mod;
int prime[N+10],cnt;
bool vis[N+10];
bool is_prime(ll x)
{
    for(int i=0;i<cnt&&(ll)prime[i]*prime[i]<=x;i++)
    {
        if(x%prime[i]==0)
            return 0;
    }
    return 1;
}
void get_prime()
{
    for(int i=2;i<=N;i++)
    {
        if(!vis[i])
            prime[cnt++]=i;
        for(int j=0;j<cnt&&(ll)i*prime[j]<=N;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0) break;    
        }    
    }    
}
ll mul(ll a,ll b)
{
    ll res=0;
    while(b)
    {
        if(b&1) res=(res+a)%mod;
        a=(a+a)%mod;
        b>>=1;
    }
    return res%mod;
}
ll poww(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=mul(res,a);
        a=mul(a,a);
        b>>=1;
    }
    return res;
}
int main(void)
{
    int t;
    get_prime(); 
    ll p,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&p);
        mod=p;
        q=p-1;
        while(!is_prime(q)) q--;
        ll ans=p-1;
        for(ll i=q+1;i<=p-1;i++)
            ans=mul(ans,poww(i,mod-2));
        printf("%lld\n",ans);
    }
    return 0;
}