An easy problem
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33073 Accepted Submission(s): 21354

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

問題連結：HDU - 2055

問題簡述：把A~ Z當做1~26，把a ~當做-1 ~-26，再加上一個數y，輸出結果，多組輸入輸出。

問題分析：‘A’的ASCII碼為65，將其減去64便可用‘A’代表1，同理，‘a’的ASCII碼為97，將其減 去96並取負數便可用‘a’表示-1，其他的‘A’~‘Z’ ‘a’ ~‘z’也同理。將其值加上y值並輸出即為所求值。

程式說明：i和n用於接下來輸入輸出的組數。ch代表字母，y代表要加的值。getchar（）用於接收每次scanf後多餘的字元‘\n’。

AC通過程式：

#include<stdio.h>

int main()
{
	int i;
	int n,y;
	char c;
	scanf("%d",&n);
	getchar();
	for(i=0;i<n;i++)
	{
		scanf("%c%d",&c,&y);
		getchar();
		if(c>='A'&&c<='Z')
		{
			printf("%d\n",y+c-64);
		}else printf("%d\n",y-c+96);
	}
	return 0;
}