while(n!=0)
{
num[i++]=n%2;
n /= 2; //类似模拟除法的过程
}
for(j=i-1;j>=0;j--)
printf("%d",num[j]);
Bitset
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6113 Accepted Submission(s): 4738
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
#include<stdio.h>
int main()
{
int num[10500];
int i,n,j;
while( scanf("%d",&n) == 1 )
{
i = 0;
while(n!=0)
{
num[i++]=n%2;
n /= 2;
}
for(j=i-1;j>=0;j--)
printf("%d",num[j]);
printf("\n");
}
return 0;
}
int main()
{
int num[10500];
int i,n,j;
while( scanf("%d",&n) == 1 )
{
i = 0;
while(n!=0)
{
num[i++]=n%2;
n /= 2;
}
for(j=i-1;j>=0;j--)
printf("%d",num[j]);
printf("\n");
}
return 0;
}