HDU6608-Fansblog（Miller_Rabbin素数判定，威尔逊定理应用，乘法逆元）

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

题意：

找出Q,Q为比P小的数中的最大素数，求Q!

题解：

用Miller_Rabbin素数检测快速找出Q,用威尔逊定理 ，求出（P-1）!

根据乘法逆元，用除的模求出Q！

Code:

 #include <iostream>
 #include <algorithm>
 #include <time.h>
 using namespace std;
 /**
 Miller_Rabin 算法进行素数测试
 快速判断一个<2^63的数是不是素数,主要是根据费马小定理
 */
 #define ll __int128
 const int S=; ///随机化算法判定次数
 ll MOD;
 ///计算ret=(a*b)%c  a,b,c<2^63
 ll mult_mod(ll a,ll b,ll c)
 {
     a%=c;
     b%=c;
     ll ret=;
     ll temp=a;
     while(b)
     {
         if(b&)
         {
             ret+=temp;
             if(ret>c)
                 ret-=c;//直接取模慢很多
         }
         temp<<=;
         if(temp>c)
             temp-=c;
         b>>=;
     }
     return ret;
 }

 ///计算ret=(a^n)%mod
 ll pow_mod(ll a,ll n,ll mod)
 {
     ll ret=;
     ll temp=a%mod;
     while(n)
     {
         if(n&)
             ret=mult_mod(ret,temp,mod);
         temp=mult_mod(temp,temp,mod);
         n>>=;
     }
     return ret;
 }

 ///通过费马小定理 a^(n-1)=1(mod n)来判断n是否为素数
 ///中间使用了二次判断,令n-1=x*2^t
 ///是合数返回true，不一定是合数返回false
 bool check(ll a,ll n,ll x,ll t)
 {
     ll ret=pow_mod(a,x,n);
     ll last=ret;//记录上一次的x
     for(int i=;i<=t;i++)
     {
         ret=mult_mod(ret,ret,n);
         if(ret==&&last!=&&last!=n-)
             return true;//二次判断为是合数
         last=ret;
     }
     if(ret!=)
         return true;//是合数,费马小定理
     return false;
 }

 ///Miller_Rabbin算法
 ///是素数返回true(可能是伪素数)，否则返回false
 bool Miller_Rabbin(ll n)
 {
     if(n<) return false;
     if(n==) return true;
     if((n&)==) return false;//偶数
     ll x=n-;
     ll t=;
     while((x&)==)
     {
         x>>=;
         t++;
     }
     srand(time(NULL));
     for(int i=;i<S;i++)
     {
         ll a=rand()%(n-)+; // 生成随机数 0<a<=n-1  去试试
         if(check(a,n,x,t))
             return false;
     }
     return true;
 }

 //------------------------------------------------------------------------求素数
 inline ll pow(const ll n, const ll k) {
     ll ans = ;
     for (ll num=n,t=k;t;num=num*num%MOD,t>>=) if(t&) ans=ans*num%MOD;
     return ans%MOD;
 }

 inline ll inv(const ll num) {
     return pow(num, MOD - );
 }
 //求乘法逆元

 int main(){
     ll ans;
     int T;
     long long a;
     cin>>T;
     while(T--){
         cin>>a;
         MOD=a;
         a--;
         while(!Miller_Rabbin(a)) a--;
         ans=MOD-;
         for(ll i=a+;i<=MOD-;i++){
             ans=(ans%MOD*inv(i)%MOD)%MOD;
         }
         a=ans;
         cout<<a<<'\n';
     }
 }