The statement of this problem is very simple: you are to determine how many squares of the chessboard can be attacked by a knight standing alone on the board. Recall that a knight moves two squares forward (horizontally or vertically in any direction) and then one square sideways (perpedicularly to the first direction).

Input

The first line contains the number N of test cases, 1 ≤ N ≤ 100. Each of the following N lines contains a test: two characters. The first character is a lowercase English letter from 'a' to 'h' and the second character is an integer from 1 to 8; they specify the rank and file of the square at which the knight is standing.

Output

Output N lines. Each line should contain the number of the squares of the chessboard that are under attack by the knight.

Sample

input output
3
a1
d4
g6
2
8
6
Problem Author: folklore Problem Source: Fifth High School Children Programming Contest, Ekaterinburg, March 02, 2002
// Ural Problem 1197. Lonesome Knight
// Verdict: Accepted
// Submission Date: 10:31:16 14 Jan 2014
// Run Time: 0.015s
//
// 版权所有(C)acutus。(mail: acutus@126.com)
// 博客:http://www.cnblogs.com/acutus/
// [解题方法]
// 简单题,直接按题意判断即可 #include<stdio.h> int countNumber(int i, int j)
{
int count = ;
if((i - ) >= ) {
if((j - ) >= ) count++;
if((j + ) <= ) count++;
}
if((i - ) >= ) {
if((j - ) >= ) count++;
if((j + ) <= ) count++;
}
if((i + ) <= ) {
if((j - ) >= ) count++;
if((j + ) <= ) count++;
}
if((i + ) <= ) {
if((j - ) >= ) count++;
if((j + ) <= ) count++;
}
return count;
} void solve()
{
int N, n;
char c;
scanf("%d", &N);
getchar();
while(N--) {
scanf("%c%d", &c, &n);
getchar();
printf("%d\n",countNumber(c - 'a' + , n));
}
} int main()
{
solve();
return ;
}

Ural 1197 - Lonesome Knight的更多相关文章

  1. LeetCode 1197. Minimum Knight Moves

    原题链接在这里:https://leetcode.com/problems/minimum-knight-moves/ 题目: In an infinite chess board with coor ...

  2. Ural 1298 Knight 题解

    目录 Ural 1298 Knight 题解 题意 题解 程序 Ural 1298 Knight 题解 题意 给定一个\(n\times n(1\le n\le8)\)的国际象棋棋盘和一个骑士(基本上 ...

  3. 【计算几何】URAL - 2101 - Knight's Shield

    Little Peter Ivanov likes to play knights. Or musketeers. Or samurai. It depends on his mood. For pa ...

  4. POJ2488A Knight's Journey[DFS]

    A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41936   Accepted: 14 ...

  5. LightOJ 1197 Help Hanzo(区间素数筛选)

    E - Help Hanzo Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit ...

  6. Knight Moves

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  7. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

  8. ural 2071. Juice Cocktails

    2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and ...

  9. ural 2073. Log Files

    2073. Log Files Time limit: 1.0 secondMemory limit: 64 MB Nikolay has decided to become the best pro ...

随机推荐

  1. linux所有信息查询网址

  2. 转: vim简明教程

    vim的学习曲线相当的大(参看各种文本编辑器的学习曲线),所以,如果你一开始看到的是一大堆VIM的命令分类,你一定会对这个编辑器失去兴趣的.下面的文章翻译自<Learn Vim Progress ...

  3. WEB和APP谁是互联网未来

    据中国多家权威报告显示,作为多年专业化互联网公司炎帝网络科技综合评估预测,预计2016年全球互联网设备将达到100亿部.如果届时全球人口达到73亿,意味着平均每人将有1.4部设备.智能交通将增长50倍 ...

  4. [Windows编程] #pragma once 和#ifndef ... #define ... #endif 比较

    C++中防止头文件被多次include 的常见方式有: 1) 用#ifndef ...  #define ... #endif  宏 #ifndef __MYHEADER_H__#define __M ...

  5. SpringMVC静态文件(图片)访问+js访问 简单小例子

    项目文件布局: web.xml文件: <?xml version="1.0" encoding="UTF-8"?> <web-app vers ...

  6. linux内核源码阅读之facebook硬盘加速利器flashcache

    从来没有写过源码阅读,这种感觉越来越强烈,虽然劣于文笔,但还是下定决心认真写一回. 源代码下载请参见上一篇flashcache之我见 http://blog.csdn.net/liumangxiong ...

  7. 讲讲金融业务(一)--自助结算终端POS

    之前在群里和大家聊天的时候,发现好多人对银行业务比較感兴趣,或许是由于大家对银行不了解,以为非常神奇的样子.全部,从这周開始我打算把我肚子里的墨水慢慢地倒出来,和大家分享分享.   在技术还不发达的时 ...

  8. SVNserver的本地搭建和使用

    Subversion是优秀的版本号控制工具,其具体的的长处和具体介绍,这里就不再多说. 首先来下载和搭建SVNserver. 如今Subversion已经迁移到apache站点上了,下载地址: htt ...

  9. Sql语句之select 5种查询

    select 5种子句:注意顺序where / group by /having / order by / limit / 清空表中的数据:truncate 表名: 导入表结构(不含数据): crea ...

  10. Stack集合、queue集合、hashtable集合

    1.栈:Stack,先进后出,一个一个赋值,一个一个取值,按顺序. .count           取集合内元素的个数 .push()         将元素一个一个推入集合中//stack集合存入 ...