hdoj1905 Pseudoprime numbers （基础数论）

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 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

题意：输入两个数p,a；如果a的p次方对p取余等于a，并且p不是素数，则输出“yes”,否则输出“no”.

这里用到快速幂求余技巧

#include <iostream>
#include <stdio.h>
using namespace std;
bool isprime(long long n){
    for (long long i = ; i*i <= n; i++){
        if (n%i == )
            return false;
    }
    return true;
}
long long qmod(long long a, long long r, long long m){
    long long res = ;
    while (r){
        if (r & )
            res = res*a%m;
        a = a*a%m;
        r >>= ;
    }
    return res;
}
int main(){
    long long p, a;
    while (scanf("%I64d%I64d", &p, &a) && p&&a){
        if (!isprime(p) && qmod(a, p, p) == a)
            printf("yes\n");
        else
            printf("no\n");
    }
    return ;
}