POJ 3641

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6044		Accepted: 2421

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

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

快速幂

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 
 6 using namespace std;
 7 
 8 typedef long long ll;
 9 
10 int p,a;
11 
12 bool judge() {
13         for(int i = 2; i * i <= p; i++) {
14                 if(p % i == 0) return false;
15         }
16 
17         return true;
18 }
19 bool  mod_pow(ll x,ll n) {
20         ll res = 1;
21         while(n > 0) {
22                 if(n & 1) res = res * x % p;
23                 x = x * x % p;
24                 n >>= 1;
25         }
26 
27         return res == a;
28 }
29 
30 int main() {
31         //freopen("sw.in","r",stdin);
32 
33         while(~scanf("%d%d",&p,&a) && p && a) {
34                 if(!judge() && mod_pow(a,p)) printf("yes
");
35                 else printf("no
");
36 
37         }
38 
39         return 0;
40 }

View Code