二进制手表

class Solution {
public:
    vector<string> readBinaryWatch(int turnedOn) {
        vector<string> ans;
        for(int h=0;h<12;h++)
        {
            for(int m=0;m<60;++m)
            {
                //__builtin_popcoun 返回二进制中‘1’的个数
                if (__builtin_popcount(h) + __builtin_popcount(m) == turnedOn) 
                {
                    ans.push_back(to_string(h)+":"+(m<10?"0":"")+to_string(m));
                }
            }
        }
        return ans;
 
    }
};