hdu 3826

Squarefree number

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1984    Accepted Submission(s): 526

Problem Description

In mathematics, a squarefree number is one which is divisible by no perfect squares, except 1. For example, 10 is square-free but 18 is not, as it is divisible by 9 = 3^2. Now you need to determine whether an integer is squarefree or not.

 

Input

The first line contains an integer T indicating the number of test cases.
For each test case, there is a single line contains an integer N.

Technical Specification

1. 1 <= T <= 20
2. 2 <= N <= 10^18

 

Output

For each test case, output the case number first. Then output "Yes" if N is squarefree, "No" otherwise.

 

Sample Input

2

30

75

 

Sample Output

Case 1: Yes

Case 2: No

 

Author

hanshuai

采用如果素因子超过2次，NO。这个很好理解。

这样还是会超时。10^18次方，最坏的情况下10^9次方。

但是，我们只需要枚举10^6内的素数就可以。为什么？

 

假如n 有(10^6,10^9)的一个素因子平方构成 ,不在10^6范围，但是可能直接开方判断即可。

还有假如是(10^,10^9)的一个素因子平方*a构成，则a必在10^6内。不然数字就超过10^18次方了。

我们用[1,10^6]内的素数，刷一遍的时候，就可能除去a，然后又可以开方，直接判断就行。

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<math.h>
using namespace std;
typedef __int64 LL;

bool s[1000002];
LL prime[78500],len;
void init()
{
    int i,j;
    len=0;
    memset(s,false,sizeof(s));
    for(i=2;i<=1000000;i++)
    {
        if(s[i]==true)continue;
        prime[++len]=i;
        for(j=i*2;j<=1000000;j=j+i)
            s[j]=true;
    }
}
LL Euler(LL n)
{
    LL i,num;
    for(i=1;prime[i]*prime[i]<=n && i<=len;i++)
    {
        if(n%prime[i]==0)
        {
            num=0;
            while(n%prime[i]==0)
            {
                n=n/prime[i];
                num++;
                if(num>=2) return 0;
            }
        }
    }
    return n;
}
int main()
{
    init();
    int T,t;
    LL n;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%I64d",&n);
        LL ans = Euler(n);
        if(ans==0)
        {
            printf("Case %d: No
",t);
        }
        else
        {
            LL cur=(LL)sqrt(ans*1.0);
            if(cur*cur==ans)printf("Case %d: No
",t);
            else printf("Case %d: Yes
",t);
        }
    }
    return 0;
}