HDOJ 2055

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

#include<stdio.h> int main() { int T,ans,d; char ch; scanf("%d", &T); while(T--) { getchar(); //添加该句接收缓存中的换行或者其它字符,以免被赋值给x ,加上fflush(stdin)亦可 scanf("%c %d", &ch, &d);//字符型也需要取地址符 if(ch >= 65 && ch <= 90) ch = ch - 64; else if(ch >= 97 && ch <= 122) ch = (ch - 96) * (-1); ans = ch + d; printf("%d\n", ans); } return 0; }