NYOJ 217 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

思路：略。

#include <stdio.h>

int main()
{
int m;
scanf("%d",&m);

while (m--)
{
char x;
int y;
getchar();
scanf("%c %d",&x,&y);
if (x<= 'Z' && x >= 'A')
{
x = x - 'A'+1;
}
else
{
x = 'a'- x -1;
}
printf("%d\n",x+y);
}
return 0;
}