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hdu-Bitset
Bitset
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12693 Accepted Submission(s): 9772
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
代码:
#include<stdio.h>
int main()
{
int n,a[100];
while(~scanf("%d",&n))
{
int i=0,j;
while(n)
{
a[i++]=n%2;
n/=2;
}
for(j=i-1;j>0;--j)
{
if(a[j]!=0)
break;
}
for(;j>=0;--j)
{
printf("%d",a[j]);
}
printf("\n");
}
return 0;
}hdu-Bitset
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