Mobile keyboard (day 73)

1： subject

 Please calculate according to the mobile phone keyboard （9  Key input method ） How to enter letters , Type the given string （ Consisting of lowercase letters ） The time it takes .

 
 The specific typing rules and time spent are described below ：

 For characters on the same key , for example  a,b,c  All in  “1”  On key , Input  a  Just press once , Input  c  You need to press it three times in a row .

 If two consecutive characters are not on the same key , You can press , Such as ：ad  Need to press twice ,kz  Need to press  6  Next .

 If two consecutive characters are on the same key , You need to wait a while between the two buttons , Such as  ac, I pressed  a  after , It will take a while to press  c.

 Now suppose it takes a period of time to press each time , The waiting time takes two time periods .

 Input format 
 Input contains multiple sets of test data .

 Each group of data occupies one row , Contains a string of lowercase letters .

 Output format 
 For each group of input , One line of output indicates the time it takes to type a given string .

 Data range 
 Each input contains at most  100  Group test data .
 All strings are no longer than  100.

 sample input ：
bob
www
 sample output ：
7
7
 difficulty ： Simple 
 when / Empty limit ：1s / 64MB
 Total number of passes ：943
 Total attempts ：1398
 source ： Tsinghua University postgraduate entrance examination machine test questions 
 Algorithm tags 

2： Code implementation

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int number[] = {
    
    2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 
    6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9
};

int main()
{
    
    string str;
    while (cin >> str)
    {
    
        int res = 0, last = -1;
        for (auto c: str)
        {
    
            int t = c - 'a';
            for (int i = t; i >= 0; i -- )
                if (number[i] == number[t])
                    res ++ ;
                else
                    break;
            if (number[t] == number[last]) res += 2;
            last = t;
        }
        cout << res << endl;
    }

    return 0;
}