hdoj2055 An easy problem

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
 
代码如下：
#include<stdio.h>#include<string.h>int f[550000]={0};
int main(){    int i,n,y;    char x,j;    for(j='a';j<='z';j++)        f[j]=f[j-1]-1;    for(j='A';j<='Z';j++)        f[j]=f[j-1]+1;    scanf("%d",&n);        while(n--)    {        getchar();        scanf("%c %d",&x,&y);        printf("%d\n",f[x]+y);    }    return 0;}