HDU 3826 Squarefree number 简单的数论题

Squarefree number

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1476    Accepted Submission(s): 390

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

23075

 

Sample Output

Case 1: YesCase 2: No

 

算法分析：

对于2-10^18的任意一个数 都能转化成如下的形式  ……等等 a，b，c，d都是质数

然后 

情况1  将输入的num 先从 2-10^6中的质数进行相除，有一个质数能连出两次以上 就输出no 

情况2  如果能整除 num=num/i 然后 如果 相除后num==1 输出yes

进行完一二操作后 本题中剩下的num这个数 （因为不存在1-10^6的质数） 最多只有两个质数 

情况3   只有两种结果 两个10^6-10^9 的质数相乘 或者只有一个质数 在10^6-10^18次方之间。  这样只要判断第一种结果是否是两个相同质数相乘的情况了~~

post code：

#include<stdio.h>#include<string.h>#include<math.h>#define MAX 1100000#define MIN 0.00001int a[MAX];int b[500000],flag;int select()   //筛选法求素数{     int i,j=1;     memset( a, 0, sizeof(a) );     a[1]=1;     for( i=2; i*i<MAX; i++ )     {          if( a[i]==0 )for( j=2*i; j<MAX; j+=i )                          {                             a[j]=1;                          }          }               j=1;     for( i=2; i<MAX ;i++)     {        if(a[i]==0){b[j]=i;j++;}          }     return j;} void issquare( double num )   //判断开方是否为整数{    double a;    a=sqrt(num);    if( a-floor(a)<MIN )flag=1;     }int main(){     int i,len,x,sum,ji=0;     __int64 num;     len=select();        scanf("%d",&x);     while(x--)     {          ji++;          flag=0;          scanf("%I64d",&num);          for( i=1; i<len; i++)          {                 sum=0;                 while( num%b[i]==0 )                      {                        num/=b[i];                        sum++;                      }                      if( sum>=2 ){                                 flag=1;                                 break;                             }          }          if(flag==1){printf("Case %d: No\n",ji);continue;}  //情况1          if(num==1){printf("Case %d: Yes\n",ji);continue;}  //情况2          issquare((double) num);                           //情况3          if(flag==1)printf("Case %d: No\n",ji);          else printf("Case %d: Yes\n",ji);                   }}