HDU - 2055——An easy problem

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

问题链接：HDU - 2055

问题简述：把A~ Z当做1~26，把a ~当做-1 ~-26，再加上一个数y，输出结果，多组输入输出。

问题分析：‘A’的ASCII码为65，将其减去64便可用‘A’代表1，同理，‘a’的ASCII码为97，将其减 去96并取负数便可用‘a’表示-1，其他的‘A’~‘Z’ ‘a’ ~‘z’也同理。将其值加上y值并输出即为所求值。

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程序说明：i和n用于接下来输入输出的组数。ch代表字母，y代表要加的值。getchar（）用于接收每次scanf后多余的字符‘\n’。

AC通过程序：

#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;
}