hdu 3826

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

题解：

 参考大神题解[题解](http://blog.csdn.net/acdreamers/article/details/8678651)

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;typedef long long LL;const int maxn=1000000;long long prime[maxn+5];long long n;void getPrime(){    memset(prime,0,sizeof(prime));    for(LL i=2;i<=maxn;i++)    {        if(!prime[i])        {            prime[++prime[0]]=i;        }        for(LL j=1;j<=prime[0]&&prime[j]<=maxn/i;j++)        {            prime[prime[j]*i]=1;            if(i%prime[j]==0)                break;        }    }}bool Test(){    int i;    for(i=1;i<prime[0]&&prime[i]<n;i++)    {        if(n%prime[i]==0)        {            n/=prime[i];            if(n%prime[i]==0)            {                //cout<<n<<" "<<" "<<i<<prime[i]<<endl;                return false;            }        }    }    return true;}int main(){    int T;    cin>>T;    int kn=1;    getPrime();    while(T--)    {       scanf("%I64d",&n);       bool flag=0;       if(!Test())            flag=true;       if (!flag)        {            double temp = sqrt(n * 1.0);            if ((int)(temp + 0.5) == temp) flag = 1;        }       if(flag) printf("Case %d: No\n",kn++);       else printf("Case %d: Yes\n",kn++);    }    return 0;}