FZU 1649 Prime number or not （大素数判定）

Your task is simple.

Give you a number N, you should judge whether N is a prime number or not.

Input

There are multiple test cases. For each test case, there is an integer N(2<=N<=10^18).

Output

For each test case, you should output whether N is a prime number or not.

If N is a prime number , you should output “It is a prime number.”; otherwise you should output “It is not a prime number.”;

Sample Input

2

4

Sample Output

It is a prime number.

It is not a prime number.

大素数判定：Miller Rabbin

费马小定理：

a为整数，n是素数，且a，n互质，则有a^(n-1)≡1(mod n) ，即：a^(n-1)模n得1。

快速判定一个数是否为素数的方法：

如果存在一个整数a，使得a^(n-1)≡1(mod n) ，则称n为基于a的伪素数，当有多个满足关系的a时，则n为素数的概率趋向于1。所以取多个a测试一下即可。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef __int64 LL;
#define maxn 50
LL sum(LL a,LL b,LL mod)         //快速乘
{
LL ans=0;
while(b)
{
if(b&1)
ans=(ans+a)%mod;
a=a*2%mod;
b/=2;
}
return ans;
}

LL pow(LL a,LL b,LL mod)       //快速幂
{
LL ans=1;
while(b)
{
if(b&1)
ans=sum(ans,a,mod);
a=sum(a,a,mod);
b/=2;
}
return ans;
}

int Miller_Rabbin(LL x)
{
if(x==2)
return 1;
if(x<2||!(x&1))
return 0;
srand((unsigned)time(NULL));          //随机数
for(int i=1;i<=maxn;i++)
{
LL a=rand()%(x-2)+1;
if(pow(a,x-1,x)!=1)       // a^(n-1) % n == 1
return 0;
}
return 1;
}

int main()
{
LL n;
while(cin>>n)
{
if(Miller_Rabbin(n))
printf("It is a prime number.\n");
else
printf("It is not a prime number.\n");
}
return 0;
}