第十届蓝桥杯决赛JavaC组-奇数倍数-递增序列-平方拆分

一.奇数倍数

请你找到最小的整数 X 同时满足：
X 是 2019 的整倍数
X 的每一位数字都是奇数
解：通过暴力循环可解

public class qsbs{
	public static void main(String[] args) {
		a: for (int i = 2019, n = 2019; true; n = i += 2019) {
			do {
				if ((n & 1) == 0)
					continue a;
			} while ((n /= 10) > 0);
			System.out.println(i);
			break;
		}
	}
}

二.递增序列
对于一个字母矩阵，我们称矩阵中的一个递增序列是指在矩阵中找到两个字母，它们在同一行，同一列，或者在同一 45 度的斜线上，这两个字母从左向右看、或者从上向下看是递增的。
例如，如下矩阵中
LANN
QIAO
有LN、LN、AN、AN、IO、AO、LQ、AI、NO、NO、AQ、IN、AN 等 13 个递增序列。注意当两个字母是从左下到右上排列时，从左向右看和从上向下看是不同的顺序。
对于下面的 30 行 50 列的矩阵，请问总共有多少个递增序列？（如果你把以下文字复制到文本文件中，请务必检查复制的内容是否与文档中的一致。在试题目录下有一个文件 inc.txt，内容与下面的文本相同）
解：通过暴力循环可解

VLPWJVVNNZSWFGHSFRBCOIJTPYNEURPIGKQGPSXUGNELGRVZAG
SDLLOVGRTWEYZKKXNKIRWGZWXWRHKXFASATDWZAPZRNHTNNGQF
ZGUGXVQDQAEAHOQEADMWWXFBXECKAVIGPTKTTQFWSWPKRPSMGA
BDGMGYHAOPPRRHKYZCMFZEDELCALTBSWNTAODXYVHQNDASUFRL
YVYWQZUTEPFSFXLTZBMBQETXGXFUEBHGMJKBPNIHMYOELYZIKH
ZYZHSLTCGNANNXTUJGBYKUOJMGOGRDPKEUGVHNZJZHDUNRERBU
XFPTZKTPVQPJEMBHNTUBSMIYEGXNWQSBZMHMDRZZMJPZQTCWLR
ZNXOKBITTPSHEXWHZXFLWEMPZTBVNKNYSHCIQRIKQHFRAYWOPG
MHJKFYYBQSDPOVJICWWGGCOZSBGLSOXOFDAADZYEOBKDDTMQPA
VIDPIGELBYMEVQLASLQRUKMXSEWGHRSFVXOMHSJWWXHIBCGVIF
GWRFRFLHAMYWYZOIQODBIHHRIIMWJWJGYPFAHZZWJKRGOISUJC
EKQKKPNEYCBWOQHTYFHHQZRLFNDOVXTWASSQWXKBIVTKTUIASK
PEKNJFIVBKOZUEPPHIWLUBFUDWPIDRJKAZVJKPBRHCRMGNMFWW
CGZAXHXPDELTACGUWBXWNNZNDQYYCIQRJCULIEBQBLLMJEUSZP
RWHHQMBIJWTQPUFNAESPZHAQARNIDUCRYQAZMNVRVZUJOZUDGS
PFGAYBDEECHUXFUZIKAXYDFWJNSAOPJYWUIEJSCORRBVQHCHMR
JNVIPVEMQSHCCAXMWEFSYIGFPIXNIDXOTXTNBCHSHUZGKXFECL
YZBAIIOTWLREPZISBGJLQDALKZUKEQMKLDIPXJEPENEIPWFDLP
HBQKWJFLSEXVILKYPNSWUZLDCRTAYUUPEITQJEITZRQMMAQNLN
DQDJGOWMBFKAIGWEAJOISPFPLULIWVVALLIIHBGEZLGRHRCKGF
LXYPCVPNUKSWCCGXEYTEBAWRLWDWNHHNNNWQNIIBUCGUJYMRYW
CZDKISKUSBPFHVGSAVJBDMNPSDKFRXVVPLVAQUGVUJEXSZFGFQ
IYIJGISUANRAXTGQLAVFMQTICKQAHLEBGHAVOVVPEXIMLFWIYI
ZIIFSOPCMAWCBPKWZBUQPQLGSNIBFADUUJJHPAIUVVNWNWKDZB
HGTEEIISFGIUEUOWXVTPJDVACYQYFQUCXOXOSSMXLZDQESHXKP
FEBZHJAGIFGXSMRDKGONGELOALLSYDVILRWAPXXBPOOSWZNEAS
VJGMAOFLGYIFLJTEKDNIWHJAABCASFMAKIENSYIZZSLRSUIPCJ
BMQGMPDRCPGWKTPLOTAINXZAAJWCPUJHPOUYWNWHZAKCDMZDSR
RRARTVHZYYCEDXJQNQAINQVDJCZCZLCQWQQIKUYMYMOVMNCBVY
ABTCRRUXVGYLZILFLOFYVWFFBZNFWDZOADRDCLIRFKBFBHMAXX

import java.io.BufferedReader;
 
public class ddxl{
	static String str = "VLPWJVVNNZSWFGHSFRBCOIJTPYNEURPIGKQGPSXUGNELGRVZAG"
			+ "SDLLOVGRTWEYZKKXNKIRWGZWXWRHKXFASATDWZAPZRNHTNNGQF"
			+ "ZGUGXVQDQAEAHOQEADMWWXFBXECKAVIGPTKTTQFWSWPKRPSMGA"
			+ "BDGMGYHAOPPRRHKYZCMFZEDELCALTBSWNTAODXYVHQNDASUFRL"
			+ "YVYWQZUTEPFSFXLTZBMBQETXGXFUEBHGMJKBPNIHMYOELYZIKH"
			+ "ZYZHSLTCGNANNXTUJGBYKUOJMGOGRDPKEUGVHNZJZHDUNRERBU"
			+ "XFPTZKTPVQPJEMBHNTUBSMIYEGXNWQSBZMHMDRZZMJPZQTCWLR"
			+ "ZNXOKBITTPSHEXWHZXFLWEMPZTBVNKNYSHCIQRIKQHFRAYWOPG"
			+ "MHJKFYYBQSDPOVJICWWGGCOZSBGLSOXOFDAADZYEOBKDDTMQPA"
			+ "VIDPIGELBYMEVQLASLQRUKMXSEWGHRSFVXOMHSJWWXHIBCGVIF"
			+ "GWRFRFLHAMYWYZOIQODBIHHRIIMWJWJGYPFAHZZWJKRGOISUJC"
			+ "EKQKKPNEYCBWOQHTYFHHQZRLFNDOVXTWASSQWXKBIVTKTUIASK"
			+ "PEKNJFIVBKOZUEPPHIWLUBFUDWPIDRJKAZVJKPBRHCRMGNMFWW"
			+ "CGZAXHXPDELTACGUWBXWNNZNDQYYCIQRJCULIEBQBLLMJEUSZP"
			+ "RWHHQMBIJWTQPUFNAESPZHAQARNIDUCRYQAZMNVRVZUJOZUDGS"
			+ "PFGAYBDEECHUXFUZIKAXYDFWJNSAOPJYWUIEJSCORRBVQHCHMR"
			+ "JNVIPVEMQSHCCAXMWEFSYIGFPIXNIDXOTXTNBCHSHUZGKXFECL"
			+ "YZBAIIOTWLREPZISBGJLQDALKZUKEQMKLDIPXJEPENEIPWFDLP"
			+ "HBQKWJFLSEXVILKYPNSWUZLDCRTAYUUPEITQJEITZRQMMAQNLN"
			+ "DQDJGOWMBFKAIGWEAJOISPFPLULIWVVALLIIHBGEZLGRHRCKGF"
			+ "LXYPCVPNUKSWCCGXEYTEBAWRLWDWNHHNNNWQNIIBUCGUJYMRYW"
			+ "CZDKISKUSBPFHVGSAVJBDMNPSDKFRXVVPLVAQUGVUJEXSZFGFQ"
			+ "IYIJGISUANRAXTGQLAVFMQTICKQAHLEBGHAVOVVPEXIMLFWIYI"
			+ "ZIIFSOPCMAWCBPKWZBUQPQLGSNIBFADUUJJHPAIUVVNWNWKDZB"
			+ "HGTEEIISFGIUEUOWXVTPJDVACYQYFQUCXOXOSSMXLZDQESHXKP"
			+ "FEBZHJAGIFGXSMRDKGONGELOALLSYDVILRWAPXXBPOOSWZNEAS"
			+ "VJGMAOFLGYIFLJTEKDNIWHJAABCASFMAKIENSYIZZSLRSUIPCJ"
			+ "BMQGMPDRCPGWKTPLOTAINXZAAJWCPUJHPOUYWNWHZAKCDMZDSR"
			+ "RRARTVHZYYCEDXJQNQAINQVDJCZCZLCQWQQIKUYMYMOVMNCBVY"
			+ "ABTCRRUXVGYLZILFLOFYVWFFBZNFWDZOADRDCLIRFKBFBHMAXX";
	static char[][] c = new char[30][50];
	static int count = 0;
 
	public static void main(String[] args) {
		for (int i = 0; i < c.length; i++) {
			for (int j = 0; j < c[i].length; j++) {
				c[i][j] = str.charAt(i * c[i].length + j);
			}
		}
		for (int i = 0; i < c.length; i++) {
			for (int j = 0; j < c[i].length; j++) {
				check(i, j);
			}
		}
		System.out.println(count);
	}
 
	public static void check(int i, int j) {
		// 右
		for (int k = 1; k < c[i].length - j; k++) {
			if (j + k < c[i].length && c[i][j] < c[i][j + k]) {
				count++;
			}
		}
		// 下
		for (int k = 1; k < c.length - i; k++) {
			if (i + k < c.length && c[i][j] < c[i + k][j]) {
				count++;
			}
		}
		// 右下角
		for (int k = 1; k < c.length - i; k++) {
			if (i + k < c.length && j + k < c[i].length && c[i][j] < c[i + k][j + k]) {
				count++;
			}
		}
		// 右上角
		for (int k = 1; k < c.length; k++) {
			if (i - k >= 0 && j + k < c[i].length && c[i][j] < c[i - k][j + k]) {
				count++;
			}
		}
		// 左下角
		for (int k = 1; k < c.length; k++) {
			if (i + k < c.length && j - k >= 0 && c[i][j] < c[i + k][j - k]) {
				count++;
			}
		}
	}
}

三、平方拆分
本题总分：10 分

问题描述

将 2019 拆分为若干个两两不同的完全平方数之和，一共有多少种不同的方法？
注意交换顺序视为同一种方法，例如 13^2 + 25^2 + 35^2 = 2019 与 13^2 + 35^2 +25^2 = 2019 视为同一种方法(^代表平方)。
解：通过暴力循环可解

public class pfcf{
	static int count = 0;
 
	public static void main(String[] args) {
		dfs(2019, -1);
		System.out.println(count);
	}
 
	public static void dfs(int num, int start) {
		if (num < 0) {
			return;
		}
		if (num == 0) {
			count++;
		} else {
			for (int i = start + 1, high = (int) Math.sqrt(num); i <= high; i++)
				dfs(num - i * i, i);
		}
	}
}