HDU 1905 Pseudoprime numbers （快速幂求余）

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本文介绍了一种使用费马小定理来判断特定整数是否为伪素数的方法，并提供了一个C++实现示例。通过快速幂求余算法，可以有效地进行大数运算，适用于在计算机科学中解决与素数测试相关的问题。

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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

快速幂求余，原理我没看懂，只能当模板用了~

#include <iostream>
using namespace std;


bool isprime(long long a) {
    for(long long i = 2; i * i <= a; i++) {
        if (a % i == 0) return false;
    }
    return true;
}

long long qmod(long long a, long long r, long long m) {
    long long res = 1;
    while (r) {
        if (r & 1) 
            res = res * a % m;
        a = a * a % m;
        r >>= 1;
    }
    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 0;
}