while(n!=0)

        {

            num[i++]=n%2;

            n /= 2; //类似模拟除法的过程

        }

        for(j=i-1;j>=0;j--)

            printf("%d",num[j]);

Bitset

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6113    Accepted Submission(s): 4738

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

 

 

 

 

#include<stdio.h>

int main()

{

    int num[10500];

    int i,n,j;

    while( scanf("%d",&n) == 1 )

    {

        i = 0;

       

while(n!=0)

        {

            num[i++]=n%2;

            n /= 2;

        }

        for(j=i-1;j>=0;j--)

            printf("%d",num[j]);

        printf("\n");

    }

      return 0;

}