C++ 递归，移位Bitset

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 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

123

 

Sample Output

11011

 题目简单，代码如下：

#include<iostream>using namespace std;void sol(int n) {if(n) sol(n>>1),cout << n%2;//利用n>>1等于n/2 ,n<<1等于n*2 else return ;}int main() {int n;while(cin >> n) {sol(n);cout<< endl;}return 0; }