Prime Test
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
2
5
10
Sample Output
Prime
2
Source
题目类型：Miller_Rabin测试和Pollard_Rho大整数分解
算法分析：先判断n是否是素数，如果是则直接输出”Prime”，反之则分解n大整数，生成n所有的素因子，然后找到最小的那一个即可，注意srand()要注销掉，否则会RE
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 7; const int maxn = 10000 + 66; long long factor[maxn], fac_len; const int testnum = 8; //随机算法判定次数，一般8~10就够了 // 计算ret = (a*b)%c a,b,c < 2^63 long long mult_mod (long long a,long long b,long long mod) { a %= mod; b %= mod; long long ans = 0; long long temp = a; while (b) { if (b & 1) { ans += temp; if (ans > mod) ans -= mod;//直接取模慢很多 } temp <<= 1; if (temp > mod) temp -= mod; b >>= 1; } return ans; } long long pow_mod (long long a,long long n,long long mod) { long long ans = 1; long long temp = a % mod; while (n) { if (n & 1) ans = mult_mod (ans, temp, mod); temp = mult_mod (temp, temp, mod); n >>= 1; } return ans; } // 通过 a^(n-1)=1(mod n)来判断n是不是素数 // n-1 = x*2^t 中间使用二次判断 // 是合数返回true, 不一定是合数返回false bool check (long long a, long long n, long long x, long long t) { long long ans = pow_mod (a, x, n); long long last = ans; for(int i = 1; i <= t; i++) { ans = mult_mod (ans, ans, n); if(ans == 1 && last != 1 && last != n - 1) return true;//合数 last = ans; } if(ans != 1) return true; else return false; } //************************************************** // Miller_Rabin算法 // 是素数返回true,(可能是伪素数) // 不是素数返回false //************************************************** bool Miller_Rabin (long long n) { if (n < 2) return false; if (n == 2) return true; if ((n&1) == 0) return false;//偶数 long long x = n - 1; long long t = 0; while ((x&1) == 0) { x >>= 1; t++; } // srand (time (NULL)); for (int i = 0; i < testnum; i++) { long long a = rand () % (n - 1) + 1; if (check (a, n, x, t)) return false; } return true; } long long gcd (long long a, long long b) { if (a == 0) return 1; if (a < 0) return gcd (-a, b); while (b) { long long t = a % b; a = b; b = t; } return a; } long long Pollard_Rho (long long x, long long c) { long long i = 1, k = 2; long long x0 = rand () % x; long long y = x0; while (1) { i++; x0 = (mult_mod (x0, x0, x) + c) % x; long long d = gcd (y - x0, x); if (d != 1 && d != x) return d; if (y == x0) return x; if (i == k) { y = x0;k += k; } } } //对n进行素因子分解 void FindFac (long long n) { if (Miller_Rabin (n))//素数 { factor[fac_len++] = n; return ; } long long p = n; while (p >= n) p = Pollard_Rho (p, rand () % (n - 1) + 1); FindFac (p); FindFac (n / p); } int main() { // freopen ("aaa.txt", "r", stdin); long long t; scanf ("%lld", &t); while (t--) { long long n; scanf ("%lld", &n); if (Miller_Rabin (n)) printf ("Prime\n"); else { fac_len = 0; FindFac (n); long long minval = INF; for (long long i = 0; i < fac_len; i++) minval = min (minval, factor[i]); printf ("%lld\n", minval); } } return 0; } |
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