hdu1905(判断质数+快速幂)

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3192    Accepted Submission(s): 1441

Problem 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 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

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

 

Author

Gordon V. Cormack

#include<stdio.h>#include<iostream>#include<math.h>using namespace std;typedef long long ll;//判断是不是素数+快速幂取模 int prime(int x){for(int i=2;i*i<x;i++){if(x%i==0)return 0;}return 1;}ll fsm(ll a,ll b,ll mod){ll res=1;while(b){if(b&1) res=res*a%mod;a=a*a%mod;b>>=1; }return res%mod;}int main(){int p,a;while(scanf("%d%d",&p,&a)!=EOF){if(!p&&!a)break;if(prime(p))printf("no\n");else{if(fsm(a,p,p)==a){printf("yes\n");}else printf("no\n");}  }  return 0;}