Bitset（十进制转二进制）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=2051

Bitset

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14434    Accepted Submission(s): 10985

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

 

Source

校庆杯Warm Up

 

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AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
using namespace std;
int main()
{
	int n,a[111],cnt,i;
	while(scanf("%d",&n)!=EOF)
	{
		cnt=0;
		while(n)
		{
			a[cnt++]=n%2;
			n/=2;
		}
		for(i=cnt-1;i>=0;i--)
		{
			printf("%d",a[i]);
		}
		printf("\n");
	}
 }