【Bitset ２０５１】

Bitset

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10109    Accepted Submission(s): 7832

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<iostream>
using namespace std;
int a[21];
int main(){
	int n;
		while(cin>>n){
		int i=0;
		while(n){
			a[i]=n%2;
			n/=2;
			i++;
		}
		for(int j=i-1;j>=0;j--)
			cout<<a[j];
		cout<<endl;
	}
}

关于求二进制：依次对２求余，然后用数组保存，从高位输出。