Canada Cup 2016 B. Food on the Plane

B. Food on the Plane

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A new airplane SuperPuperJet has an infinite number of rows, numbered with positive integers starting with 1 from cockpit to tail. There are six seats in each row, denoted with letters from 'a' to 'f'. Seats 'a', 'b' and 'c' are located to the left of an aisle (if one looks in the direction of the cockpit), while seats 'd', 'e' and 'f' are located to the right. Seats 'a' and 'f' are located near the windows, while seats 'c' and 'd' are located near the aisle.

 

It's lunch time and two flight attendants have just started to serve food. They move from the first rows to the tail, always maintaining a distance of two rows from each other because of the food trolley. Thus, at the beginning the first attendant serves row 1 while the second attendant serves row 3. When both rows are done they move one row forward: the first attendant serves row 2 while the second attendant serves row 4. Then they move three rows forward and the first attendant serves row 5 while the second attendant serves row7. Then they move one row forward again and so on.

Flight attendants work with the same speed: it takes exactly 1 second to serve one passenger and 1 second to move one row forward. Each attendant first serves the passengers on the seats to the right of the aisle and then serves passengers on the seats to the left of the aisle (if one looks in the direction of the cockpit). Moreover, they always serve passengers in order from the window to the aisle. Thus, the first passenger to receive food in each row is located in seat 'f', and the last one — in seat 'c'. Assume that all seats are occupied.

Vasya has seat s in row n and wants to know how many seconds will pass before he gets his lunch.

Input

The only line of input contains a description of Vasya's seat in the format ns, where n (1 ≤ n ≤ 1018) is the index of the row and s is the seat in this row, denoted as letter from 'a' to 'f'. The index of the row and the seat are not separated by a space.

Output

Print one integer — the number of seconds Vasya has to wait until he gets his lunch.

Examples

input

1f

output

1

input

2d

output

10

input

4a

output

11

input

5e

output

18

Note

In the first sample, the first flight attendant serves Vasya first, so Vasya gets his lunch after 1 second.

In the second sample, the flight attendants will spend 6 seconds to serve everyone in the rows 1 and 3, then they will move one row forward in 1 second. As they first serve seats located to the right of the aisle in order from window to aisle, Vasya has to wait 3 more seconds. The total is 6 + 1 + 3 = 10.

思路：每四个一组，每一组用时16秒，然后每一组的偶数加7，奇数加0，然后再加上座位数。

代码：

#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#include<math.h>#include<stdlib.h>#include<stack>#include<vector>#include<string.h>#include<map>#define INF 0x3f3f3f3f3fusing namespace std;int main(){    long long n;    char c;    while(~scanf("%I64d%c",&n,&c))    {        getchar();        long long sum=((n-1)/4)*16;//看有多少组，注意不要加上本组因为本组不完全所以（n-1）        int t=n%4;看看剩下组不成一组的有几个        if(t%2==0)//如果是偶数的话就加上7        {            sum+=7;        }        if(c=='f')//加上座位数            sum+=1;        else if(c=='e')            sum+=2;        else if(c=='d')            sum+=3;        else if(c=='a')            sum+=4;        else if(c=='b')            sum+=5;        else if(c=='c')            sum+=6;        printf("%I64d\n",sum);//输出结果    }}

以上是参考别人的代码写出的优化代码，下面是本人的超级麻烦代码：

#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#include<math.h>#include<stdlib.h>#include<stack>#include<vector>#include<string.h>#include<map>#define INF 0x3f3f3f3f3fusing namespace std;int main(){  long long flag;  char a;   while(~scanf("%I64d%c",&flag,&a))   {       getchar();       long long sum,tt,t;       if(flag%2==1&&(flag-1)%4==0)       {           t=1;           sum=flag/2;       }       if(flag%2==1&&(flag-3)%4==0)       {           t=0;           sum=(flag-2)/2;       }       if(flag%2==0&&(flag-2)%4==0)       {           tt=1;          sum=flag/2;       }       if(flag%2==0&&(flag-4)%4==0)       {           tt=0;           sum=(flag-2)/2;       }      //printf("%d\n",sum);      if((flag%2==0&&tt==1)||(flag%2==1&&t==1))      {        sum=(sum*6+flag-1);      }      else      {          sum=(sum*6+flag-3);      }      //printf("%d\n",sum);       if(a<='f'&&a>='d')       {           sum+=('f'-a+1);       }       else       {           sum+=3;           sum+=(a-'a'+1);       }       printf("%I64d\n",sum);   }}

真的超级麻烦但是当时的第一反应就是这个方法，然后一旦有一个方法就陷进去了，一直在修改代码，然后就形成了这个超级麻烦的代码，所以还是想法太麻烦，刷题量不够呀。