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>void main(){ int n; while(scanf("%d",&n)!=EOF) { for(int i=0;i<n;i++) { getchar(); char c; int k; scanf("%c%d",&c,&k); if(c>='a'&&c<='z') printf("%dn",96+k-(int)c); if(c>='A'&&c<='Z') printf("%dn",k+(int)c-64); //printf("%dn",k+z); } }}乍一看这题目的名字，感觉又是迷惑性战术。嗯。看完题目，好像标题的还真没骗人。
哦事实证明，我确实花了很多时间解这题。主要是少了那句getchar() 因为上一次输入的换行符还存在缓冲区里（但是不加这句前好像感觉没影响输出结果这是为啥。）