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


[align=left]Problem Description[/align]
Give you a number on base ten,you should output it on
base two.(0 < n < 1000)

[align=left]Input[/align]
For each case there is a postive number n on base ten,
end of file.

[align=left]Output[/align]
For each case output a number on base two.

[align=left]Sample Input[/align]

1

2

3

[align=left]Sample Output[/align]

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;
}