Pseudoprime numbers【poj-3641】【快速幂】

Pseudoprime numbers Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10620 Accepted: 4578

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

题意：输入两个数，判断公式 a^p=a(mod p)是否成立，即a的p次方对p求模是否等于a。只有当p是非素数时且等式成立时，才输出yes。其他情况输出no。

题解：利用快速幂算法，简单的素数判定。

代码如下：

#include<cstdio> typedef long long ll; bool Prime_judge(ll q) { for(ll i=2;i*i<=q;i++) { if(q%i==0) return false; } return true; } ll p,a; ll mod_pow(ll x,ll n){ ll res=1; while(n>0){ if(n&1) res=res*x%p; x=x*x%p; n>>=1; } return res; } int main() { while(scanf("%lld%lld",&p,&a)){ if(p==0&&a==0) break; if(Prime_judge(p)){ printf("no\n"); }else{ ll ans=mod_pow(a,p); ans=ans%p; if(ans==a) printf("yes\n"); else printf("no\n"); } } return 0; }