poj 3641 Pseudoprime numbers（快速幂）

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = 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  < p ≤  and  < 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

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

感觉好久没A题了，脑子都快生锈了，所有赶紧做做题。

求(a^p)%p==a,数据大所有用long long

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define N 1000000
 #define inf 1e12
 ll pow_mod(ll a,ll n,ll MOD)
 {
     if(n==)
        return %MOD;
     ll tt=pow_mod(a,n>>,MOD);
     ll ans=tt*tt%MOD;
     if(n&)
       ans=ans*a%MOD;
     return ans;
 }
 int main()
 {
    ll p,a;
    while(scanf("%I64d%I64d",&p,&a)==){
       if(p== && a==){
          break;
       }
       int flag=;
       for(int i=;i<(int)sqrt(p+0.5);i++){
          if(p%i==){
             flag=;
             break;
          }
       }
       if(flag==){
          printf("no\n");
          continue;
       }
       ll ans=pow_mod(a,p,p);

       //printf("%I64d\n",ans);
       if(ans==a){
          printf("yes\n");
       }else{
          printf("no\n");
       }
    }
     return ;
 }