POJ 1811 Prime Test（Miller-Rabin & Pollard-rho素数测试）

Description

Given a big integer number, you are required to find out whether it's a prime number.
Input

The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N < 254).
Output

For each test case, if N is a prime number, output a line containing the word "Prime", otherwise, output a line containing the smallest prime factor of N.

题目大意：给一个数n，问是不是素数，若是输出Prime，若不是输出其最小的非1因子。
思路：http://www.2cto.com/kf/201310/249381.html
模板盗自：http://vfleaking.blog.163.com/blog/static/1748076342013231104455989/

代码（375MS）：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iterator>
#include <vector>
using namespace std;

typedef long long LL;

LL modplus(LL a, LL b, LL mod) {
LL res = a + b;
return res < mod ? res : res - mod;
}

LL modminus(LL a, LL b, LL mod) {
LL res = a - b;
return res >= 0 ? res : res + mod;
}

LL mult(LL x, LL p, LL mod) {
LL res = 0;
while(p) {
if(p & 1) res = modplus(res, x, mod);
x = modplus(x, x, mod);
p >>= 1;
}
return res;
}

LL power(LL x, LL p, LL mod) {
LL res = 1;
while(p) {
if(p & 1) res = mult(res, x, mod);
x = mult(x, x, mod);
p >>= 1;
}
return res;
}

bool witness(LL n, LL p) {
int t = __builtin_ctz(n - 1);
LL x = power(p % n, (n - 1) >> t, n), last;
while(t--) {
last = x, x = mult(x, x, n);
if(x == 1 && last != 1 && last != n - 1) return false;
}
return x == 1;
}

const int prime_n = 5;
int prime[prime_n] = {2, 3, 7, 61, 24251};

bool isPrime(LL n) {
if(n == 1) return false;
if(find(prime, prime + prime_n, n) != prime + prime_n) return true;
if(n % 2 == 0) return false;
for(int i = 0; i < prime_n; i++)
if(!witness(n, prime[i])) return false;
return true;
}

LL getDivisor(LL n) {
int c = 1;
while (true) {
int i = 1, k = 2;
LL x1 = 1, x2 = 1;
while(true) {
x1 = modplus(mult(x1, x1, n), c, n);
LL d = __gcd(modminus(x1, x2, n), n);
if(d != 1 && d != n) return d;
if(x1 == x2) break;
i++;
if(i == k) x2 = x1, k <<= 1;
}
c++;
}
}

void getFactor(LL n, vector<LL> &ans) {
if(isPrime(n)) return ans.push_back(n);
LL d = getDivisor(n);
getFactor(d, ans);
getFactor(n / d, ans);
}

int main() {
int T; LL n;
scanf("%d", &T);
while(scanf("%I64d", &n) != EOF) {
if(isPrime(n)) puts("Prime");
else {
vector<LL> ans;
getFactor(n, ans);
printf("%I64d\n", *min_element(ans.begin(), ans.end()));
}
}
}

View Code