南阳OJ-题目217：a letter and a number

最新推荐文章于 2022-07-21 23:04:47 发布

原创最新推荐文章于 2022-07-21 23:04:47 发布 · 343 阅读

0 ·

CC 4.0 BY-SA版权

本文为博主原创文章，未经博主允许不得转载。

文章标签：

#java编程 #算法

南阳OJ 语言入门专栏收录该内容

21 篇文章

订阅专栏

题目描术：

a letter and a number

时间限制：3000 ms | 内存限制：65535 KB

难度：1

描述

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).

输入

On the first line, contains a number T(0<T<=10000).then T lines follow, each line is a case.each case contains a letter x and a number y(0<=y<1000).

输出

for each case, you should the result of y+f(x) on a line

样例输入

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

样例输出

2.运行代码

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*
 * Dragon 2017.8.31
 * a letter and a number
 */
public class Num_217 {

	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(in.readLine());
		
		while(t-->0)
		{
			String[] input=in.readLine().split(" ");
			int letter=input[0].charAt(0);
			int number=Integer.parseInt(input[1]);
			
			if(letter>=97 && letter<=122)
				System.out.println(number-letter+96);
			else
				System.out.println(number+letter-64);
		}
	}

}