NYOJ-a letter and a number

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

代码:

#include<stdio.h>
int main()
{
int t,y;
char ch;
scanf("%d",&t);
while(t--)
{
getchar();  //注意字符输入时可能出现的问题,否则ch可能会接收'\n';
scanf("%c%d",&ch,&y);
if(ch>='a'&&ch<='z')
printf("%d\n",y-(ch-'a'+1));
else
printf("%d\n",y+(ch-'A'+1));
}
return 0;
}