利用变换次数

解题思路:利用左支不变右支变

注意事项:

参考代码:

#include <bits/stdc++.h>
using namespace std;
int main(){
    int m;
    cin>>m;
    for(int z=0;z<m;z++){
        int nn;
        long long kk;
        cin>>nn>>kk;
        int change=0;
        long long currentk=kk;
        if(nn==1){
            cout<<"RED"<<endl;
            continue;
        }
        for(int i=1;i<nn;i++){
            if(currentk%2==1){
                currentk=(currentk+1)/2;
            }
            else {
                currentk=currentk/2;
                change+=1;
            }
        }
        if(change%2==0){
            cout<<"RED"<<endl;
        }
        else cout<<"BLACK"<<endl;
    }
}