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

代码:

[objc] view plaincopyprint?

1. #include<stdio.h>  
2. int main() 
3. { 
4.     int t,y; 
5.     char ch; 
6.     scanf("%d",&t); 
7.     while(t--) 
8.     { 
9.         getchar();  //注意字符输入时可能出现的问题,否则ch可能会接收‘\n‘; 
10.         scanf("%c%d",&ch,&y); 
11.         if(ch>=‘a‘&&ch<=‘z‘) 
12.         printf("%d\n",y-(ch-‘a‘+1)); 
13.         else 
14.         printf("%d\n",y+(ch-‘A‘+1)); 
15.     } 
16.     return 0; 
17. } 

NYOJ-a letter and a number