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

十进制转化为二进制  太水了

#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\n",sum);
	}
	return 0;
}