hdoj 1905 Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2446    Accepted Submission(s): 1000

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
快速幂取模：
题意：两个数，q,a.,如果q是素数，输出no；如果(a^q)%q==a,输出yes否则输出no!
代码：
#include<stdio.h>bool judge(__int64 n)//判断素数 {__int64 i;for(i=2;i*i<=n;i++)if(n%i==0)return false;return true;}int  Pseudoprime(__int64 a,__int64 b,__int64 k)//判断是否符合条件 {__int64 ans=1;    a=a%k;while(b>0){if(b%2)ans=(ans*a)%k;b=b/2;a=(a*a)%k;}return ans;}int main(){   __int64 n,m,i;   while(scanf("%I64d%I64d",&n,&m),n,m)   {     if(judge(n))     { printf("no\n");     continue;  } __int64 s=Pseudoprime(m,n,n); if(s==m) printf("yes\n"); else printf("no\n");   }   return 0;}