[POJ 3641] Pseudoprime numbers【快速幂】

  • 2017-12-30
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

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

Waterloo Local Contest, 2007.9.23

Solution:

又一道大水题。

O(sqrt(n)) 素数判定 + 快速幂模板。

Code: O(Tsqrt(p)) [160K, 16MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

ll p, a;

inline bool isPrime(ll num){
	for(register ll i = 2; i * i <= num; i++) if(num % i == 0) return 0;
	return 1;
}

inline ll fastpow(ll bas, ll ex, ll md){
	ll iden = bas, res = 1;
	while(ex){
		if(ex & 1) res = res * iden % md;
		ex >>= 1, iden = iden * iden % md;
	}
	return res;
}

int main(){
	while(scanf("%lld%lld", &p, &a) != EOF && p > 2){
		if(!isPrime(p) && fastpow(a, p, p) == a) puts("yes");
		else puts("no");
	}
	return 0;
}

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OPEN AT 2017.12.10

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