Prime Test
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 37752 | Accepted: 10155 | |
| 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
模板代码：
/*来自：李双智学长模板*/
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<stdlib.h>
#include<time.h>
#define times 20
using namespace std;
long long total;
long long factor[110];
long long qmul(long long a,long long b,long long M){
a%=M;
b%=M;
long long ans=0;
while (b){
if (b&1){
ans=(ans+a)%M;
}
a=(a<<=1)%M;
b>>=1;
}
return ans%M;
}//快乘，因为两个longlong的数相乘可能会溢出，所以这里转乘法为加法，思想和快速幂相似
long long qpow(long long a,long long b,long long int M){
long long ans=1;
long long k=a;
while(b){
if(b&1)ans=qmul(ans,k,M)%M;
k=qmul(k,k,M)%M;
b>>=1;
}
return ans%M;
}//快速幂
bool witness(long long a,long long n,long long x,long long sum){
long long judge=qpow(a,x,n);
if (judge==n-1||judge==1)return 1;
while (sum--){
judge=qmul(judge,judge,n);
if (judge==n-1)return 1;
}
return 0;
}
bool miller(long long n){
if (n<2)return 0;
if (n==2)return 1;
if ((n&1)==0)return 0;
long long x=n-1;
long long sum=0;
while (x%2==0){
x>>=1;
sum++;
}
for (long long i=1;i<=times;i++){
long long a=rand()%(n-1)+1;
if (!witness(a,n,x,sum))return 0;//费马小定理的随机数检验
}
return 1;
}//判断一个数是否为素数
long long gcd(long long a,long long b){
return b==0?a:gcd(b,a%b);
}//欧几里得算法
long long pollard(long long n,long long c){
long long x,y,d,i=1,k=2;
x=rand()%n;
y=x;
while (1){
i++;
x=(qmul(x,x,n)+c)%n;
d=gcd(y-x,n);
if (d<0)d=-d;
if (d>1&&d<n)return d;
if (y==x)return n;
if (i==k){
y=x;
k<<=1;
}
}
}
void find(long long n){
if (miller(n)){
factor[++total]=n;
return ;
}
long long p=n;
while (p>=n){
p=pollard(p,rand()%(n-1)+1);
}
find(n/p);
find(p);
}//寻找这个数的素因子，并存起来
int main(){
long long n,m,i,t;
scanf("%lld",&t);
while (t--){
scanf("%lld",&n);
if (miller(n)){
printf("Prime\n");
}
else {
memset(factor,0,sizeof(factor));
total=0;
find(n);
sort(factor+1,factor+total+1);
printf("%lld\n",factor[1]);//输出最小的素数因子
}
}
}