题解 | #手机键盘#

//判断两个相邻字母是否在同一组按键上有其他方式
//若两相邻字母相减之差（0~3）==两相邻字母按键时间之差，则需要加上等待时间
//不过我的方法也很好
#include <iostream>
using namespace std;
int a[9][5]={
{0,0,0,0,0},
{0,1,2,3,0},
{0,1,2,3,0},

{0,1,2,3,0},
{0,1,2,3,0},
{0,1,2,3,0},

{0,1,2,3,4},
{0,1,2,3,0},
{0,1,2,3,4}
};
int b[9][5]={
{0,0,0,0,0},
{0,1,2,3,0},
{0,4,5,6,0},

{0,7,8,9,0},
{0,10,11,12,0},
{0,13,14,15,0},

{0,16,17,18,19},
{0,20,21,22,0},
{0,23,24,25,26}
};

int main() {
    string st;
    
    while (cin >> st ) { 
        int sum=0;
        int len = st.length();
        int t1=0,t2=0;
        for(int i=0;i<=len-1;i++)
        {    t2=t1;//左边的字母
             t1=st[i]-'a'+1;//右边的
             
           for(int j=1;j<=8;j++)
           {
            for(int k=1;k<=4;k++)
            {
                if(t1==b[j][k])sum+=a[j][k];
                
            }
           }

           if(t2!=0)
           {
                 int x,y;
                for(int j=1;j<=8;j++)
           {
            for(int k=1;k<=4;k++)
            {   
               
                if(t1==b[j][k])x=j;
                if(t2==b[j][k])y=j;
            }
           }
                if(x==y)
                {
                    sum+=2;
                }
           }
        }
      cout<<sum<<endl;
    }
}