Miller_rabin算法+Pollard_rho算法 POJ 1811 Prime Test

POJ 1811 Prime Test

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 32534		Accepted: 8557
Case 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

2
5
10

Sample Output

Prime
2

 /*遇上一个题目不同，这个题目是输出最小的质因子*/
 #include<iostream>
 using namespace std;
 #include<cstdio>
 #define S 10
 #include<cstdlib>
 #include<ctime>
 #define ll long long
 ll cas, maxz;
 ll read()
 {
     ll ans=;char c;
     c=getchar();
     while(c<''||c>'') c=getchar();
     while(c>=''&&c<='')
     {
         ans=ans*+c-'';
         c=getchar();
     }
     return ans;
 }
 ll quick_mul_mod(ll a,ll b,ll c)//a*b%c
 {
     ll ret=;
     a%=c;b%=c;
     while(b)
     {
         if(b&)
         {
             ret+=a;
             ret%=c;
             b--;
         }
         a<<=;
         a%=c;
         b>>=;
     }
     return ret;
 }
 ll gcd(ll a,ll b)
 {
     if(a==) return ;
     if(a<) return gcd(-a,b);
     if(b==)
     return a;
     return gcd(b,a%b);
 }
 ll Pollard_rho(ll x,ll c)
 {
     ll x1=rand()%(x-)+;
     ll x2=x1;
     int i=,k=;
     while()
     {
         i++;
         x1=(quick_mul_mod(x1,x1,x)+c)%x;
         ll d=gcd(x2-x1,x);
         if(d!=&&d!=x) return d;
         if(x2==x1) return x;
         if(i==k)
         {
             x2=x1;
             k+=k;
         }
     }

 }
 ll quick_mod(ll a,ll b,ll c)//ji suan a^b%c
 {
     ll ans=;
     a%=c;
     while(b)
     {
         if(b&)
         {
             b--;
             ans=quick_mul_mod(ans,a,c);
         }
         b>>=;
         a=quick_mul_mod(a,a,c);
     }
     return ans;
 }
 bool Miller_rabin(ll n)
 {
     if(n==) return true;
     if(n<=||!(n&)) return false;
     ll u=n-,t=;
     while(!(u&))
     {
         u>>=;
         t++;
     }
     for(int i=;i<S;++i)
     {
         ll x=rand()%(n-)+;
         x=quick_mod(x,u,n);
         for(int i=;i<=t;++i)
         {
             ll y=quick_mul_mod(x,x,n);
             if(y==&&x!=&&x!=n-)
               return false;
             x=y;
         }
         if(x!=) return false;
     }
     return true;
 }
 void findpri(ll n)
 {
     if(n<=) return;
     if(Miller_rabin(n))
     {
         if(n!=)
         maxz=min(maxz,n);
         return;
     }
     ll p=n;
     while(p==n)
       p=Pollard_rho(p,rand()%(n-)+);
     findpri(p);
     findpri(n/p);
 }
 int main()
 {
     srand(time());
     cas=read();
     while(cas--)
     {
         maxz=(<<)-;/*不知道为什么这个赋初值的最大值，只有赋值为小于等于(1<<31)-1才对，我的maxz明明是long long型的啊*/
         ll n=read();
         findpri(n);
         if(Miller_rabin(n))
           printf("Prime\n");
         else printf("%lld\n",maxz);
     }
     return ;
  }