poj 3641 Pseudoprime numbers

题目连接

http://poj.org/problem?id=3641  

Pseudoprime numbers

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

快速幂。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<set>
using std::min;
using std::sort;
using std::pair;
using std::swap;
using std::vector;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1 << 17;
const int INF = ~0u >> 1;
typedef unsigned long long ull;
bool isPrime(ull n) {
    for(int i = 2; (ull)i * i <= n; i++ ) {
        if(n % i == 0) {
            return false;
        }
    }
    return n != 1;
}
ull mod_pow(ull a, ull p) {
    ull ans = 1, M = p;
    while(p) {
        if(p & 1) ans = ans * a % M;
        a = a * a % M;
        p >>= 1;
    }
    return ans;
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    ull a, p;
    while(~scanf("%lld %lld", &p, &a), a + p) {
        if(isPrime(p)) { puts("no"); continue; }
        puts(a % p == mod_pow(a, p) ? "yes" : "no");
    }
    return 0;
}