hdu2055

An easy problem

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18849    Accepted Submission(s): 12585

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
   
   
   
   

 

遇到的问题和思路：

       我还以为和杭电的2054的题目一样是大数据呢，没想到是水题。。。不过个人感觉杭电的2054的题目有点问题啊，看到discuss里面以后，我用两个人AC的代码测试了两个不同的数据，一个输出是YES 一个是NO，但是竟然都AC了。

给出代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define inf 1e-12


using namespace std;


int f[30];
char a;
int b;


int main(){
int t,indexx;
for(int i = 0;i < 29;i++)f[i] = i;
scanf("%d",&t);
getchar();
while(t--){
scanf("%c%d",&a,&b);
getchar();
if(a>='a'&&a<='z'){
indexx = a - 'a' + 1;
indexx*=-1;
}
else {
indexx = a - 'A' + 1;
}
printf("%d\n",indexx + b);
}
return 0;
}