HDU-2055 An easy problem

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

6R 1P 2G 3r 1p 2g 3

Sample Output

191810-17-14-4
#include<iostream>
using namespace std;
int main()
{
    int m,n,b[1000],i;
    char ch;
    b['a']=-1;
    for(i='b';i<='z';i++)
    b[i]=b[i-1]-1;
    cin>>n;
    while(n--)
    {
        cin>>ch>>m; 
        if(ch>='a'&&ch<='z')
        cout<<b[ch]+m<<endl;
        else 
        cout<<-b[ch+32]+m<<endl;
    }
    return 0;
}
 
这道题提交的时候一直一直报错在DEV上运行没有一点错误，一提交就WA。。。最后最后把多余定义的一个变量和一个数组删除之后居然过了。。。看来不能在HDU上提交时定义多余的变量，注意细节。

HDU-2055 An easy problem

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