An easy problem

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16516 Accepted Submission(s): 11096

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 

代码：

import java.util.*;public class Main{		public static void M1(char str,int kk){		char aa[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};												for(int i=0;i<26;i++){				if(aa[i]==str)				System.out.println((i+1)+kk);		}	}			public static void M2(char str,int kk){		char aa[]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};												for(int i=0;i<26;i++){				if(aa[i]==str)				System.out.println(-(i+1)+kk);		}	}			public static void main(String[] args) {		Scanner cin=new Scanner(System.in);		int nn=cin.nextInt();		for(int ii=0;ii
    
     ='a'&&str<='z')								M2(str,kk);			else if(str>='A'&&str<='Z')				M1(str,kk);		}	}}

本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5127031.html，如需转载请自行联系原作者