HDOJ 2055 An easy problem

简介: HDOJ 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

6

R 1

P 2

G 3

r 1

p 2

g 3


Sample Output

19

18

10

-17

-14

-4


题意: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)..

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String lowerCase = "0abcdefghijklmnopqrstuvwxyz";
        String capital = "1ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        int n =sc.nextInt();
        //sc.next();
        while(n-->0){
            String strs = sc.next();
            //System.out.println(strs);
            int y = sc.nextInt();
            char x = strs.charAt(0);
            //strs = strs.substring(2);
            //int y = Integer.parseInt(strs,10);
            //System.out.println(x);
            //System.out.println(y);
            int g=0,h=0;
            for(int i=1;i<lowerCase.length();i++){
                if(x==lowerCase.charAt(i)){
                    g=-i;
                    break;
                }
            }
            for(int i=1;i<capital.length();i++){
                if(x==capital.charAt(i)){
                    h=i;
                    break;
                }
            }
            System.out.println(y+g+h);
        }
    }
}
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