| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6644 | Accepted: 2696 |
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
#include <stdio.h>
#include <string.h>
#include <math.h>
typedef long long LL;
bool isPrime(int n) {
if(n < 2) return false;
int t = sqrt((double)n);
for(int i = 2; i <= t; ++i)
if(n % i == 0) return false;
return true;
}
LL mod_power(LL x, LL n, LL mod) {
LL ret = 1;
for( ; n > 0; n >>= 1) {
if(n & 1) ret = ret * x % mod;
x = x * x % mod;
}
return ret;
}
int main() {
LL p, a;
while(scanf("%lld%lld", &p, &a), p | a) {
if(isPrime(p)) printf("no\n");
else printf(mod_power(a, p, p) == a ? "yes\n" : "no\n");
}
return 0;
}POJ3641 Pseudoprime numbers 【快速幂】
原文：http://blog.csdn.net/chang_mu/article/details/41173473