POJ 3641 Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8846 Accepted: 3723

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

Source

Waterloo Local Contest, 2007.9.23

题意and题解：若a^p%p=a,而且a不是素数，则输出yes，所以如果a是素数或者a不是素数但是却不满足那个式子就输出no.由于此题数很大，所以要用到快速幂的方法。

<span style="font-family:SimSun;font-size:18px;">#include<cstdio>__int64 fun(__int64 a,__int64 p){__int64 c=p,ans=1,base=a;while(p){if(p&1){ans=(ans*base)%c;  //同余定理，否则可能会因数值太大而越界 }base=(base*base)%c;p=p/2;}return ans;}__int64 gcd(__int64 a){if(a%2==0)return 0;for(__int64 i=3;i*i<a;i+=2){if(a%i==0)return 0;}return 1;}int main(){__int64 p,a;while(~scanf("%I64d%I64d",&p,&a)&&!(a==0&&p==0)){if(gcd(p))   //若a是素数 {printf("no\n");continue;}else if(a==fun(a,p)){printf("yes\n");continue;}elseprintf("no\n");}return 0;}</span><span style="font-size:14px;font-family: 'Times New Roman', Times, serif;"></span>