C + + recursive, shift bitset__c++

Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 23901 accepted Submission (s): 17756


Problem Description give you in the base ten,you should output it on base two. (0 < n < 1000)
Input for each case there are 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
The topic is simple, the code is as follows:

#include <iostream>
using namespace std;

void Sol (int n) {
	if (n) sol (n>>1), cout << n%2;//use n>>1 equals N/2, n<<1 equals n*2 
	else return;

int main () {
	int n;
	while (CIN >> N) {
		sol (n);
		cout<< Endl;
	}
	return 0;