hdoj 2051 Bitset

Bitset
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14685    Accepted Submission(s): 11173
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
 
10進法をバイナリに変換 水が多すぎる
 

#include<stdio.h>

#include<math.h>

int main()

{

	int n,j,sum;

	while(scanf("%d",&n)!=EOF)

	{

		sum=0;j=0;

		while(n!=0)

		{

			sum=sum+((n%2)*pow(10,j++));

			n=n/2;

		}

		printf("%d
",sum);

	}

	return 0;

}