POJ 3641 Pseudoprime numbers

Pseudoprime numbers

Description

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 ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 2

10 3

341 2

341 3

1105 2

1105 3

0 0

Sample Output

no

no

yes

no

yes

yes

Source

​​Waterloo Local Contest​​, 2007.9.23

算法分析：

题意：

输入p，a，如果p是素数输出no，如果p不是素数，判断a^p%p==a是否成立，如果成立输出yes,否则输出no

实现：

题目不难，题意难理解而已，素数判定（不达表）+快速幂取模

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;
typedef long long ll;
ll quick_qow(ll a,ll b,ll k)
{
    long long  res=1;
    while(b>0)
    {
        if(b%2) res=(res*a)%k;
        a=(a*a)%k;
        b/=2;
    }
    return res;
}
int isPrime(int n)
{   //返回1表示判断为质数，0为非质数，在此没有进行输入异常检测
    float n_sqrt;
    if(n==2 || n==3) return 1;
    if(n%6!=1 && n%6!=5) return 0;
    n_sqrt=floor(sqrt((float)n));
    for(int i=5;i<=n_sqrt;i+=6)
    {
        if(n%(i)==0 | n%(i+2)==0) return 0;
    }
        return 1;
} 

int main()
{
     int p,a;
     while(scanf("%d%d",&p,&a)!=EOF&&(p!=0&&a!=0))
     {
        if(isPrime(p)==1 )  {cout<<"no"<<endl;continue;}
        if(quick_qow(a,p,p)==a)  cout<<"yes"<<endl;
        else  cout<<"no"<<endl;
     }
}