poj3641(Pseudoprime numbers)素性检测+快速幂
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
题意：求判断p是否是基于a的伪素数，即p不是质数且a^p ≡ a mod p
题解：米勒罗宾素性检测，二分幂取模
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <numeric>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
ll prime[6] = {2, 3, 5, 233, 331};
ll pow_mod(ll a, ll n, ll mod)
{
ll ret = 1;
while (n)
{
if (n&1)
ret = ret * a % mod;
a = a * a % mod;
n >>= 1;
}
return ret;
}
int isPrime(ll n)
{
if (n < 2 || (n != 2 && !(n&1)))
return 0;
ll s = n - 1;
while (!(s&1))
s >>= 1;
for (int i = 0; i < 5; ++i)
{
if (n == prime[i])
return 1;
ll t = s, m = pow_mod(prime[i], s, n);
while (t != n-1 && m != 1 && m != n-1)
{
m = m * m % n;
t <<= 1;
}
if (m != n-1 && !(t&1))
return 0;
}
return 1;
}
int main()
{
ll p, a;
while (~scanf("%lld%lld", &p, &a) && (p || a))
if (!isPrime(p) && pow_mod(a, p, p) == a)
printf("yes\n");
else
printf("no\n");
return 0;
}
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