1408140917-hd-Bitset.cpp

                                              Bitset

                                                 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                      Total Submission(s): 12512    Accepted Submission(s): 9628

Problem Description

Give you a number on base ten,you should output it on base two.(0 < n < 1000)

 

Input

For each case there is a postive number n on base ten, end of file.

 

Output

For each case output a number on base two.

 

Sample Input

   
   
   
   

    
    
    
    1 2 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1 10 11
   
   
   
   

 

题目大意

      进制转换问题。将给出的10进制转换为2进制，文件以EOF结束。

 

解题思路

       循环相除取余。

 

代码

#include<stdio.h>
int s[1000];
int main()
{
    int n;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        i=0;
        while(n)
        {
            s[i]=n%2;
            n/=2;
            i++;
        }
        for(j=i-1;j>=0;j--)
            printf("%d",s[j]);
        printf("\n");
    }
    return 0;
}