POJ - 2447 RSA (PollarRho大整数的因数分解)

Problem 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

2 5 10

Sample Output

Prime 2

//修改一下模板算法即可，在下面的find函数里稍加改变
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cmath>
#include<cstring>
#include<cstdlib>
#define C   240
#define N 5500
#define times 10
#define LL long long
using namespace std;
LL ct,cnt;
LL fac[N],num[N];
LL ans;
LL gcd(LL a,LL b)
{
    return b?gcd(b,a%b):a;
}//最小公约数
LL multi(LL a,LL b,LL m)
{
    LL ans=0;
    a%=m;
    while(b){
        if(b&1)
        {
            ans=(ans+a)%m;
            b--;
        }
        b>>=1;
        a=(a+a)%m;
    }
    return ans;
}
LL quick_mod(LL a,LL b,LL m)
{
    LL ans=1;
    a%=m;
    while(b)
    {
        if(b&1)
        {
            ans=multi(ans,a,m);
            b--;
        }
        b>>=1;
        a=multi(a,a,m);
    }
    return ans;
}
bool Miller_Rabin(LL n)
{
    if(n==2)return true;
    if(n<2||!(n&1)) return false;
    LL m=n-1;
    int k=0;
    while((m&1)==0)
    {
        k++;
        m>>=1;
    }
    for(int i=0;i<times;i++)
    {
        LL a=rand()%(n-1)+1;
        LL x=quick_mod(a,m,n);
        LL y=0;
        for(int j=0;j<k;j++)
        {
            y=multi(x,x,n);
            if(y==1&&x!=1&&x!=n-1)return false;
            x=y;
        }
        if(y!=1)return false;
    }
    return true;
}
LL pollard_rho(LL n,LL c)
{
    LL i=1,k=2;
    LL x=rand()%(n-1)+1;
    LL y=x;
    while(true)
    {
        i++;
        x=(multi(x,x,n)+c)%n;
        LL d=gcd((y-x+n)%n,n);
        if(1<d&&d<n)return d;
        if(y==x)return n;
        if(i==k)
        {
            y=x;
            k<<=1;
        }
    }
}
void find(LL n,LL c)
{
    if(n==1)return;
    if(Miller_Rabin(n))
    {
        ans=min(ans,n);
        return;
    }
    LL p=n;
    LL k=c;
    while(p>=n)p=pollard_rho(p,c--);
    find(p,k);
    find(n/p,k);
}
int main()
{
    LL n;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cin>>n;
        if(Miller_Rabin(n))
            printf("Prime\n");
        else
        {
            ans=n;
            find(n,120);
            cout<<ans<<endl;
        }
    }
    return 0;
}