HDOJ--2055

An easy problem

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18344    Accepted Submission(s): 12224

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

 

简单不想说什么。

以下是我的AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f(char x)
{
    if(x>='A' && x<='Z')
    return x-64;
    if(x>='a' && x<='z')
    return -x+96;
}
int main()
{
    
    int num;
    cin>>num;
    while(num--)
    {
        int y;
        char x;
        cin>>x>>y;
        int result=f(x);
        cout<<result+y<<endl;
    }
 }