Hdu3826 squarefree number

Question link:

Portal

Question:

Squarefree number Time Limit: 10000/3000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2047 accepted submission (s): 540


Problem descriptionin mathematics, a squarefree number is one which is divisible by no perfect squares, limit T 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.
Inputthe 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
Outputfor each test case, output the case number first. Then output "yes" if n is squarefree, "no" otherwise.
Sample Input

23075

 
Sample output

Case 1: YesCase 2: No

 
Authorhanshuai
Sourcethe 6th Central China Invitational programming contest and 9th Wuhan University programming contest final
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Because the range of a number is 10 to the power of 18, the factor must be less than 10 to the power of 6, then n * n> 10 to the power of 18, therefore, create a 1000000 prime number table,
The first is the prime number table, which is used to screen the prime number table. The complexity is O (ologn), which is the fastest at present ..
If a number is equal to or greater than 10 to the power of 6 after the total division of prime number 1000000, it is then multiplied after the square is opened. For more information, see section 178.

There is a very detailed explanation.

The portal is very good ..

My code is as follows:

#include<cstdio>#include<cstring>#include<cmath>const int maxn=80000+10;const int  n=1000000;int prime[maxn],vis[n];__int64 N;int  pos;int init(){    int c=0;    int m;    memset(vis,0,sizeof(vis));    m=sqrt(n+0.5);    for(int i=2;i<=m;i++)    {       if(!vis[i])       {           for(int j=i*i;j<=n;j+=i)              vis[j]=1;       }    }    for(int i=2;i<=n;i++)    {        if(!vis[i])            prime[c++]=i;    }    return c;}int judge(){   for(int i=0;i<pos;i++)      {        while(N%prime[i]==0)        {            N=N/prime[i];            if(N%prime[i]==0)              return 0;        }      }      return 1;}int main(){    __int64 x;    int i,t,cas=1,ok;    pos=init();    scanf("%d",&t);    while(t--)    {        ok=1;        scanf("%I64d",&N);        if(!judge())            ok=0;        if(N>1000000)        {            x=(int)sqrt(double(N));            if(x*x==N)                ok=0;        }        if(ok)            printf("Case %d: Yes\n",cas++);        else            printf("Case %d: No\n",cas++);    }    return 0;}