POJ 1811 Prime Test

Prime Test

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 15434 Accepted: 2909Case Time Limit: 4000MS

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.

Sample Input

2510

Sample Output

Prime2

Source

POJ Monthly

 

/*Miller Robin素性测试 + Pollard rho寻找素因子Miller Robin 和 Pollard rho的理论想非常强，细节这里就不说了，可以参考算法导论第31章*/#include <iostream>#include <ctime>#include <cmath>#define MAX_L 64 //最长位数#define TIMES 8 //miller robin素性测试的测试次数#define MAX_VAL (pow(2.0, 60)) //定义最大值#define CVAL 200using namespace std;//最小的素因子__int64 minFactor;//(1)计算a * b mod n, 思路: 利用b的二进制表示进行拆分计算//(2)例如: b = 1011101那么a * b mod n = (a * 1000000 mod n + a * 10000 mod n + a * 1000 mod n + a * 100 mod n + a * 1 mod n) mod n//(3)思路就是上面描述的那样, 那么可以用从低位往高位遍历b, 并用a来记录当前位为1的值，每次遇到b当前位为//1就将结果值加上a并 mod n，然后a 要乘以2__int64 multAndMod(__int64 a, __int64 b, __int64 n){a = a % n;__int64 res = 0;while(b){//当前位为1if(b & 1){//加上当前权位值res += a;//相当于mod nif(res >= n) res -= n;}//乘以2，提高一位a = a<<1;//mod nif(a >= n) a -= n;b = b >> 1;}return res;}//(1)计算a ^ b mod n, 思路: 和上面类似，也是利用b的二进制表示进行拆分计算//(2)例如: b = 1011101那么a ^ b mod n = [(a ^ 1000000 mod n) * (a ^ 10000 mod n) * (a ^ 1000 mod n) * (a ^ 100 mod n) * (a ^ 1 mod n)] mod n//(3)思路就是上面描述的那样, 那么可以用从低位往高位遍历b, 并用a来记录当前位为1的值，每次遇到b当前位为//1就将结果乘上a并 mod n，然后a 要乘以a以提升一位__int64 modAndExp(__int64 a, __int64 b, __int64 n){a = a % n;__int64 res = 1;while(b >= 1){//遇到当前位为1，则让res * 当前a并mod nif(b & 1)res = multAndMod(res, a, n);//a * a以提升一位a = multAndMod(a, a, n);b = b >> 1;}return res;}//MillerRobin素性测试，true:素数，flase:合数bool millerRobin(__int64 a, __int64 n){__int64 u = 0, cur = n - 1;int t = 0;bool find1 = false;while(cur != 0){if(!find1){int pb = cur % 2;if(pb == 0) t++;else find1 = true;}if(find1)break;cur = cur / 2;}u = cur;cur = modAndExp(a, u, n);__int64 now;for(int p = 1; p <= t; p++){now = modAndExp(cur, 2, n);if(cur != 1 && now == 1 && cur != n - 1){//printf("%d %d/n", cur, now);return false;}cur = now;}if(cur != 1){//printf("a:%I64d u:%I64d n:%I64d val:%I64d/n", a, u, n, start);return false;}//printf("a:%I64d u:%I64d n:%I64d val:%I64d/n", a, u, n, start);return true;}//利用Miller Robin对n进行n次素性测试bool testPrime(int times, __int64 n){if(n == 2) return true;if(n % 2 == 0) return false;__int64 a; int t;srand(time(NULL));for(t = 1; t <= times; t++){a = rand() % (n - 1) + 1;if(!millerRobin(a, n)) return false;}return true;}__int64 gcd(__int64 a, __int64 b){if(b == 0) return (a);return gcd(b, a % b);}__int64 PollardRho(__int64 n, int c){int i = 1;srand(time(NULL));__int64 x = rand() % n;__int64 y = x;int k = 2;while(true){i = i + 1;x = (modAndExp(x, 2, n) + c) % n;__int64 d = gcd(y - x, n);if(1 < d && d < n) return d;if(y == x) return n; //重复了, 说明当前x下无解，需要重新启动PollardRhoif(i == k){y = x;k = k * 2;}}}void getSmallest(__int64 n, int c){if(n == 1) return;//判断当前因子是否为素数if(testPrime(TIMES, n)){if(n < minFactor) minFactor = n;return;}__int64 val = n;//循环，知道找到一个因子while(val == n)val = PollardRho(n, c--);//二分getSmallest(val, c);getSmallest(n / val, c);}int main(){int caseN;__int64 n;scanf("%d", &caseN);while(caseN--){scanf("%I64d", &n);minFactor = MAX_VAL;if(testPrime(TIMES, n)) printf("Prime/n");else{getSmallest(n, CVAL);printf("%I64d/n", minFactor);}}return 0;}