HDU2055An easy problem

An easy problem
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5908    Accepted Submission(s): 4106


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


Author
8600

#include <stdio.h>

//算法思路:同时接受一个数字和一个字符清况,要用getchar()接受回车换行符
int main()
{
	int t,y, ans;
	char x;
	scanf("%d", &t);
	while(t--)
	{
		getchar();
		ans = 0;
		scanf("%c%d", &x, &y);
		if(x >= 'a' && x <= 'z')
		{
			ans = y - (x - 'a' + 1);
		}else
		{
			ans = y + (x - 'A' + 1);
		}
		printf("%d\n", ans);
	}
	return 0;
}