内存:128  时间:1

题目描述

Fermat’s theorem states that for any prime number p and for any integer a > 1, ap == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1,000,000,000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a. For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

输入

输出

样例输入

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

样例输出

no
no
yes
no
yes
yes

提示

代码如下

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

int f(long long n){
	int k=sqrt(n);
	int i;
	if(n==1)return 0;
	for(i=2;i<=k;i++){
		if(n%i==0)return 0;
	}
	return 1;
}

int main(){
	long long a,b;
	long long a2,b2;
	long long ans;
	while(~scanf("%lld%lld",&b,&a)){
		a2=a;
		b2=b;
		if(a==0&&b==0) break;
		if(f(b))printf("no\n");//如果b是素数,输出no
		else{//运用快速幂
			ans=1;
			while(b>0){
				if(b&1)//判断是否为奇数,相当于 if(b%2==1)
					ans=(ans*a)%b2;
				a=(a*a)%b2;
				b=b>>1;//二进制向右移一位,相当于 b=b/2;
			}
			if(ans%b2==a2)printf("yes\n");
			else printf("no\n");
		}
	}
	return 0;
}

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