LV1 previous life archives

At a glance , Use a binary tree , This feature is very consistent

This question is really not difficult compared with the lottery question , The title is concise and clear

#include<iostream>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    while(y--)
    {
        int s=1,sum=0;
        for(int i=0;i<x;i++)
        {
            char c;
            cin>>c;
            if(c=='y')
            {
               s=s*2;
            }
            else
            {
                s=s*2+1;
            }
            sum=sum*2+1;
        }
        cout<<s-sum<<endl;
    }
    return 0;
}

This kind of funny fortune telling applet is often encountered in the online world , The implementation principle is very simple , Just design a few questions , Select a path in the judgment tree according to the player's answer to each question （ As shown in the figure below ）, The conclusion is the node corresponding to the end of the path .

Now let's number the conclusions from left to right , Number from 1 Start . It is assumed that the answers here are simple “ yes ” or “ no ”, Suppose you answer “ yes ” Corresponding to the left path , answer “ no ” Corresponding to the path to the right . A series of answers for a given player , Please return the number of the conclusion .

Input format ：

The first line of input gives two positive integers ：N（≤30） The number of questions to answer in a test for players ;M（≤100） For the number of players .

And then  M  That's ok , Each line gives the player's  N  answer . Here we use  y  representative “ yes ”, use  n  representative “ no ”.

Output format ：

For each player , Output the number of its corresponding conclusion in one line .

sample input ：

3 4
yny
nyy
nyn
yyn

sample output ：

3
5
6
2