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;

}