An easy problem

主要思想在与：B - 'A' + 1 = 2 ;   'a' - b - 1 = -2;

Problem Description

we define f(A) = 1, f(a) = -1, f(B) = 2, f(b) = -2, ... f(Z) = 26, f(z) = -26;
Give you a letter x and a number y , you should output the result of y+f(x).

 

Input

On the first line, contains a number T.then T lines follow, each line is a case.each case contains a letter and a number.

 

Output

for each case, you should the result of y+f(x) on a line.

 

Sample Input

6 R 1 P 2 G 3 r 1 p 2 g 3

 

Sample Output

19 18 10 -17 -14 -4

 

 

 

#include<stdio.h>
#include<string.h>
int main()
{
    char ch[3];
     int a, n;
     scanf( "%d", &a );
     while( a-- )
     {
         int i = 0, sum = 0;
         scanf( "%s%d", ch, &n );
         if( ch[0] >= 'A' && ch[0] <= 'Z' )
                i = ch[0] - 'A' + 1;
         else if( ch[0] >= 'a' && ch[0] <= 'z' )
                i = 'a' - ch[0] - 1;
         sum = i + n ;
         printf( "%d\n", sum );
     }
     return 0;
}