HDU 1905（搜索题，逐次平方法）

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 405    Accepted Submission(s): 157

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
 
#include <iostream>#include <math.h>using namespace std;int p,a,cnt;__int64 q[40];int judge(int p){int x=(int)sqrt((double)p);int i;for(i=2;i<=x;i++){if(p%i==0)return 0;}return 1;}int pf(int n){cnt=0;__int64 sum=a;int i;if(n%2){q[cnt++]=a;n--;}while(n>0){if(n%2){q[cnt++]=sum;n--;}sum=(sum*sum)%p;if(n==2)n=0;elsen/=2;}for(i=0;i<cnt;i++)sum=(sum*q[i])%p;return (int)sum;}int main(){while(scanf("%d%d",&p,&a),p+a){if(judge(p)){printf("no/n");continue;}if(pf(p)==a)printf("yes/n");elseprintf("no/n");}return 0;}