poj 3641解题报告

 

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3611 Accepted: 1285

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th 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

3 210 3341 2341 31105 21105 30 0

Sample Output

nonoyesnoyesyes

题目大意是这样的，输入p，a，两个数如果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

代码：

语言：c++

#include<iostream>#include<cmath>using namespace std;int main(){long long a,p;int isprime(long long n);long long modular(long long a,long long r,long long m);while(cin>>p>>a){if(p==0&&a==0)break;long long result;if(isprime(p))cout<<"no"<<endl;else{if(a==modular(a,p,p))cout<<"yes"<<endl;elsecout<<"no"<<endl;}}return 0;}int isprime(long long n){if(n==2)return 1;if(n<=1||n%2==0)return 0;long long j=3;while(j<=(long long)sqrt(double(n))){if(n%j==0)return 0;j+=2;}return 1;}long long modular(long long a,long long r,long long m){long long d=1,t=a;while(r>0){if(r%2==1)d=(d*t)%m;r/=2;t=t*t%m;}return d;}