问题 F: Fansblog(威尔逊定理+Miller-Rabin素数检测算法)

题目描述
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 )

输入
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)

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

样例输入
复制样例数据
1
1000000007
样例输出
328400734
思路：威尔逊定理+Miller-Rabin素数检测算法
(P为素数)(p-2)! = 1(modp)

q!(q+1)(q+2)----- * p = 1(modp)
那么用1/(q+1)*(q+2)----- * p不就是答案了吗，除法要用逆元

#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;

ll ModMul(ll a,ll b,ll n)//快速积取模 a*b%n
{
    ll ans=0;
    while(b){
        if(b&1)
            ans=(ans+a)%n;
        a=(a+a)%n;
        b>>=1;
    }
    return ans;
}
ll ModExp(ll a,ll b,ll n)//快速幂取模 a^b%n
{
    ll ans=1;
    while(b){
        if(b&1)
            ans=ModMul(ans,a,n);
        a=ModMul(a,a,n);
        b>>=1;
    }
    return ans;
}
bool miller_rabin(ll n)//Miller-Rabin素数检测算法
{
    ll i,j,a,x,y,t,u,s=10;
    if(n==2)
        return true;
    if(n<2||!(n&1))
        return false;
    for(t=0,u=n-1;!(u&1);t++,u>>=1);//n-1=u*2^t
    for(i=0;i<s;i++){
        a=rand()%(n-1)+1;
        x=ModExp(a,u,n);
        for(j=0;j<t;j++){
            y=ModMul(x,x,n);
            if(y==1&&x!=1&&x!=n-1)
                return false;
            x=y;
        }
        if(x!=1)
       	    return false;
    }
    return true;
}
ll inv(ll a,ll mod){//求逆元 
    return ModExp(a,mod-2,mod);
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		ll n,k;
		scanf("%lld",&n);
		for(k = n-1;;k--){
			if(miller_rabin(k)){
				break;
			}
		}
		ll ans = 1;
		for(k++;k <= n-2;k++){
			ans = ModMul(ans,inv(k,n),n);
		}
		printf("%lld\n",ans);
	}
	return 0;
}

标签：prime,Miller,ll,number,Fansblog,素数,ans,he,Rabin
来源： https://blog.csdn.net/weixin_42868863/article/details/100065709