Bitset hdoj 2051

/*
Bitset
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12378    Accepted Submission(s): 9527




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
 


Author
8600 && xhd


*/




/*从数学上看，左移1位等于乘以2，右移1位等于除以2，然后再取整，移位溢出的丢弃。*/


#include<cstdio>
#include<string.h>
#include<stdlib.h>
#define Max 1000000
char a[Max];
int main()
{
  int n;
  while(~scanf("%d",&n))
  {
 int i=0;
while(n)
  {
 
  if(n&1)
  a[i++]='1';
  else
   a[i++]='0';
  n>>=1;
  }
  while(i--)
   {
printf("%c",a[i]);
   }
   printf("\n");
  }
  return 0;
}