Pseudoprime numbers（数学 + 快速幂）

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6053		Accepted: 2423

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

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（2 ~ 1000000000），a （1 ~ p）。如果p是素数则输出no，如果不是，则判断 a ^ p % p 是否等于 a，若是则输出yes，不是则输出no。

 

    思路：

    快速幂。其时间复杂度是 O(log₂N)， 与朴素的O(N)相比效率有了极大的提高。但是要变成边模，以免爆long long。

 

    AC：

#include <stdio.h>
#include <string.h>
typedef long long ll;

int pri(int a) {
    for(int i = 2;i * i <= a;i++)
        if(!(a % i)) return 0;
    return 1;
}

int mod_pow(ll x,ll n,ll mod) {
    ll ans = x,res = 1;

    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }

    return (res == ans ? 1 : 0);
}

int main() {
    int a,p;

    while(~scanf("%d%d",&p,&a) && (a + p)) {
        if(!pri(p) && mod_pow(a,p,p))    puts("yes");
        else    puts("no");
    }

    return 0;
}