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:

[objc] view plaincopyprint?

1. #include<stdio.h>
2. int main()
3. {
4. int t,y;
5. char ch;
6. scanf("%d",&t);
7. while(t--)
8. {
9. getchar();  //注意字符输入时可能出现的问题,否则ch可能会接收‘\n‘;
10. scanf("%c%d",&ch,&y);
11. if(ch>=‘a‘&&ch<=‘z‘)
12. printf("%d\n",y-(ch-‘a‘+1));
13. else
14. printf("%d\n",y+(ch-‘A‘+1));
15. }
16. return 0;
17. }