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

Sample Output

no no yes no yes yes 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; else cout<<"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; }