hdu3826 Squarefree number

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題目連結：

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題目：

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, 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. 
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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因為數的範圍為10的18次方，那麼它的因子必須是小於10的6次方的，則n*n*n>10的18次方，所以打一個1000000的素數表，
首先是素數表，用篩法打素數表。複雜度為O(ologn)，應該是目前來說最快的吧。。
如果一個數在整除素數1000000後任然大於10的6次方的話，則將其開方後再乘。詳情參見小白178面。

有個非常詳細的講解的。

傳送門  講的非常好。。

My Code如下：

#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;}