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  • Ural 1197

    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

    inputoutput
    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 = 0;
        if((i - 1) >= 1) {
            if((j - 2) >= 1) count++;
            if((j + 2) <= 8) count++;
        }
        if((i - 2) >= 1) {
            if((j - 1) >= 1) count++;
            if((j + 1) <= 8) count++;
        }
        if((i + 1) <= 8) {
            if((j - 2) >= 1) count++;
            if((j + 2) <= 8) count++;
        }
        if((i + 2) <= 8) {
            if((j - 1) >= 1) count++;
            if((j + 1) <= 8) count++;
        }
        return count;
    }
    
    void solve()
    {
        int N, n;
        char c;
        scanf("%d", &N);
        getchar();
        while(N--) {
            scanf("%c%d", &c, &n);
            getchar();
            printf("%d
    ",countNumber(c - 'a' + 1, n));
        }
    }
    
    int main()
    {
        solve();
        return 0;
    }
    作者:acutus
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
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  • 原文地址:https://www.cnblogs.com/acutus/p/3518791.html
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