Powers of Two（题目链接）
利用递归，返回每次最小的个数。
#include <iostream>
using namespace std;
int f(int n){
int sum = 1;
while(n > sum){//寻找恰比n大的2^k的数
sum *= 2;
}
if(sum == n)//若相等，则可以直接用2^k表示
return 1;
int t1 = f(n - sum/2);//加上下一个数
int t2 = f(sum - n);//减去下一个数
if(t1 > t2)
return t2 + 1;
return t1 + 1;
}
int main(){
int n ;
cin >> n;
cout << f(n) << endl;
}
Powers of Two（题目链接） 思路 利用递归，返回每次最小的个数。 代码
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