/*a letter and a number
时间限制:3000 ms | 内存限制:65535 KB
难度:1
描述
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).
输入
On the first line, contains a number T(0<T<=10000).then T lines follow, each line is a case.each case contains a letter x and a number y(0<=y<1000).
输出
for each case, you should the result of y+f(x) on a line
样例输入
6
R 1
P 2
G 3
r 1
p 2
g 3
样例输出
19
18
10
-17
-14
-4*/
#include<stdio.h>
int main()
{
int a;
scanf("%d",&a);
while(a--)
{
int m,n,k,b;
char i;
getchar();
scanf("%c",&i);
scanf("%d",&n);
if(i>64&&i<91)
printf("%d
",i-64+n);
else if(i>96&&i<123)
printf("%d
",-(i-96)+n);
}
return 0;
}