2019hdu多校 Fansblog

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

Solution:

通过严谨的计算(打表)可以发现，对于一个质数(p)，有((p-1)!\,mod\,p=p-1)

并且我们知道对于一个质数(p)，在(log \,p)次内可以找到比它小的最大的质数

那么我们每次只要判断当前数是不是质数，再把它除掉，也就是乘上它的逆元，就行了

Code:

#include<bits/stdc++.h>
#define int long long 
using namespace std;
int read(){
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
int isprime(int x){
	if(x<2) return 0;
	for(int i=2;i*i*1ll<=x;i++)
		if(x%i==0) return 0;
	return 1;
}
int dqy(int a,int b,int p){
	int ans=0;
	while(b){
		if(b&1)ans=(ans+a)%p;
		a=(a+a)%p;
		b>>=1;
	}
	return ans;
}
int quick(int a,int b,int p){
	int re=1;
	while(b){
		if(b&1) re=dqy(re,a,p);
		a=dqy(a,a,p);b>>=1;
	}return re%p;
}
void solve(){
	int p=read(),ans=p-1;
	for(int x=p-1;x;x--){
		if(isprime(x)) break;
		ans=dqy(ans,quick(x,p-2,p),p);
	}printf("%lld
",ans);
}
signed main(){
	int t=read();
	while(t--) solve();
	return 0;
}