poj 1811解题报告

 

Prime Test

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 17747 Accepted: 3644Case 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

这道题就是miiler_rabin素数测试和pollard_rho质因子分解的标准模板题。。。第一次做这样的题，这道题我是参照网上别人的代码写的，作为学习，当做模板学习了.....

代码：

语言：c++

#include<ctime>#include<iostream>using namespace std;long long factor[100],fac_top = -1;//计算两个数的gcdlong long gcd(long long a,long long b){if(a==0)return b;long long c;while(b!=0){c=b;b=a%b;a=c;}return a;}//ret = (a*b)%n (n<2^62)long long muti_mod(long long a,long long b,long long n){ long long exp = a%n, res = 0; while(b){ if(b&1){ res += exp; if(res>n) res -= n; } exp <<= 1; if(exp>n) exp -= n; b>>=1; } return res;}// ret = (a^b)%nlong long mod_exp(long long a,long long p,long long m){ long long exp=a%m, res=1; // while(p>1) { if(p&1)// res=muti_mod(res,exp,m); exp = muti_mod(exp,exp,m); p>>=1; } return muti_mod(res,exp,m);}//miller-rabin法测试素数, time 测试次数bool miller_rabin(long long n, long long times){ if(n==2)return 1; if(n<2||!(n&1))return 0; long long a, u=n-1, x, y; int t=0; while(u%2==0){ t++; u/=2; } srand(time(0)); for(int i=0;i<times;i++){ a = rand() % (n-1) + 1; x = mod_exp(a, u, n); for(int j=0;j<t;j++){ y = muti_mod(x, x, n); if ( y == 1 && x != 1 && x != n-1 ) return false; //must not x = y; } if( y!=1) return false; } return true;}long long pollard_rho(long long n,int c){//找出一个因子 long long x,y,d,i = 1,k = 2; srand(time(0)); x = rand()%(n-1)+1; y = x; while(true) { i++; x = (muti_mod(x,x,n) + c) % n; d = gcd(y-x, n); if(1 < d && d < n)return d; if( y == x) return n; if(i == k) { y = x; k <<= 1; } }}void findFactor(long long n,int k){//二分找出所有质因子，存入factor if(n==1)return; if(miller_rabin(n, 10)){factor[++fac_top] = n;return;}long long p = n;while(p >= n)p = pollard_rho(p,k--);//k值变化，防止死循环findFactor(p,k);findFactor(n/p,k);}int main(){ long long t,n,min; cin>>t; while(t--){ cin>>n; fac_top = min = -1; if(miller_rabin(n,10)) cout<<"Prime"<<endl;else{ findFactor(n,107); for(int i=0;i<=fac_top;i++){ if(min<0||factor[i]<min) min = factor[i];} cout<<min<<endl; } } return 0;}