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("%d\n",96+k-(int)c);
if(c>='A'&&c<='Z')
printf("%d\n",k+(int)c-64);
//printf("%d\n",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。)