SOLUTION INFO
C++ · main.cpp
#include<bits/stdc++.h>
using namespace std;
const string qwerty[3] = {"qwertyuiop", "asdfghjkl", "zxcvbnm"};
const string OnlyLeft = "qwertasdfgzxcv";
int X[33], Y[33];
bool isLeft(char x){ return OnlyLeft.find(x) != string::npos; }
int dist(char a, char b){return abs(Y[a-'a']-Y[b-'a']) + abs(X[a-'a']-X[b-'a']); }
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
for(int i=0;i<3;i++){
for(int j=0;j<(int)qwerty[i].size();j++){
Y[qwerty[i][j] - 'a'] = i;
X[qwerty[i][j] - 'a'] = j;
}
}
char L, R; cin >> L >> R;
string word; cin >> word;
int answer = 0;
for(auto &w: word){
if(isLeft(w)) {
answer += dist(L, w) + 1;
L = w;
}
else {
answer += dist(R, w) + 1;
R = w;
}
}
cout << answer;
return 0;
}
SOLUTION DESCRIPTION
풀이 설명
등록된 풀이 설명이 없습니다.