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 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

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==0)
       return 1%MOD;
    ll tt=pow_mod(a,n>>1,MOD);
    ll ans=tt*tt%MOD;
    if(n&1)
      ans=ans*a%MOD;
    return ans;
}
int main()
{
   ll p,a;
   while(scanf("%I64d%I64d",&p,&a)==2){
      if(p==0 && a==0){
         break;
      }
      int flag=0;
      for(int i=2;i<(int)sqrt(p+0.5);i++){
         if(p%i==0){
            flag=1;
            break;
         }
      }
      if(flag==0){
         printf("no
");
         continue;
      }
      ll ans=pow_mod(a,p,p);

      //printf("%I64d
",ans);
      if(ans==a){
         printf("yes
");
      }else{
         printf("no
");
      }
   }
    return 0;
}