<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[qq_26793545的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/qq_26793545</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; qq_26793545]]></copyright><item><title><![CDATA[快速渡河 贪心]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104742444</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104742444</guid><author>qq_26793545</author><pubDate>Sun, 08 Mar 2020 22:31:23 +0800</pubDate><description><![CDATA[题目：
每个人过河都有自己的过河时间，有n个人想过河，但只有一只小船，最多只能装2个人，每一次过河，过河时间为用时最多的那人过河时间，如果还有人没有过河，那么过去一个用时最少的送回船。问n人过河最少要多少时间    ？

import java.util.Arrays;
import java.util.Scanner;
/*
* 题解：
*     明显用贪心算法。主要就是分析情况，并做出最优解...]]></description><category></category></item><item><title><![CDATA[dfs 素数环问题]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104719846</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104719846</guid><author>qq_26793545</author><pubDate>Sat, 07 Mar 2020 19:36:02 +0800</pubDate><description><![CDATA[素数环问题：
输入一个正整数n,对前n个数进行排序并要求两数相邻之和为素数。
解题思想：
当然用全排列也可以，但最优的还是是用dfs一步步的去找，并剪枝。
package com.lanqiaobei;

import java.util.Scanner;
//遍历前n项的相邻数的和，并输出
public class 素数环dfs {
    private static int n;
    p...]]></description><category></category></item><item><title><![CDATA[蓝桥杯 JAVA 数独dfs]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104710610</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104710610</guid><author>qq_26793545</author><pubDate>Sat, 07 Mar 2020 10:38:39 +0800</pubDate><description><![CDATA[就是通过深度搜索一步步的去找正确的结果
package com.lanqiaobei;


import java.util.Scanner;

public class 数独dfs {
    //打印符合条件的结果
    public  static void print(char[][] table){
        for(int i=0;i&lt;9;i++)
           ...]]></description><category></category></item><item><title><![CDATA[n皇后问题-递归]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104702114</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104702114</guid><author>qq_26793545</author><pubDate>Fri, 06 Mar 2020 19:00:16 +0800</pubDate><description><![CDATA[递归回溯的简单的应用
package com.lanqiaobei;

import java.util.Scanner;

public class n皇后dfs {
    private static int n;
    private static int count;
    //检查当前位置是否可用
    public static boolean check(int[] a,in...]]></description><category></category></item><item><title><![CDATA[Java中十六进制与十进制得转化]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104314707</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104314707</guid><author>qq_26793545</author><pubDate>Fri, 14 Feb 2020 16:36:39 +0800</pubDate><description><![CDATA[// 十进制转化为十六进制，结果为C8。

Integer.toHexString(200);


// 十六进制转化为十进制，结果140。
Integer.parseInt("8C",16);


]]></description><category></category></item><item><title><![CDATA[递归的应用]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104289065</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104289065</guid><author>qq_26793545</author><pubDate>Wed, 12 Feb 2020 23:34:20 +0800</pubDate><description><![CDATA[求最大子序列问题
/*

求最大子序列问题；通过递归只能解决一些小规模的问题；
思想：
①也是一样把问题分开一步一步的来求。
②分别把两个字串的第一个字母拿出来，进行比较。分别得到两种情况；
③然后分别对得到的情况进行递归；
④再处理边界情况，也就是跳出的条件；
*/

 public class Maxpub {
	
public static void main(String[] args) ...]]></description><category></category></item><item><title><![CDATA[递归的应用]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104289054</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104289054</guid><author>qq_26793545</author><pubDate>Wed, 12 Feb 2020 23:32:46 +0800</pubDate><description><![CDATA[从M个球中取出N的球的又多少中的去发
/*

不放回的分球法；
就是通过构建出一个分开的区域，进行递归
*/

public class fenqiu {
public static void main(String[] args) {
	int t=f(5,2);
	System.out.println(t);
}
public static int f(int m,int n){
	if(m&...]]></description><category></category></item><item><title><![CDATA[递归]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104289012</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104289012</guid><author>qq_26793545</author><pubDate>Wed, 12 Feb 2020 23:30:37 +0800</pubDate><description><![CDATA[求n个元素的全队列
/*

全队列问题 就是求n个元素的全队列
思想：
①先创建一个数组，转化成对这个数组的元素进行全队列打印；
②拆分先偷懒解决一个问题（首元素的位子进行遍历排序），把剩下的交给下一个人；
③遍历排序每一种的情况，最后用完后一定记得回溯的原先的状态；
*/

public class quanpailie {
public static void main(String[] ar...]]></description><category></category></item><item><title><![CDATA[二叉树的深度遍历算法]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/104143035</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/104143035</guid><author>qq_26793545</author><pubDate>Sun, 02 Feb 2020 11:56:19 +0800</pubDate><description><![CDATA[小球下落  ：
完全二叉树，最大深度D。所有节点从上到下从左到右编号为1,2,3…2的D次方-1.在节点1处放一个小球，它会往下落。每个内节点上都由一个开关，初始全部关闭，当每次有小球落到一个开关上时，它的状态都会改变。当小球达到一个内节点时，如果该节点上的开关关闭，往左走，否则往右走，直到遍历到叶子节点。
问题，输入 深度和小球个数
输出节点编号
思路
直接遍历每一个小球经过二叉树的情况，并记录...]]></description><category></category></item><item><title><![CDATA[数学解法]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/99655027</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/99655027</guid><author>qq_26793545</author><pubDate>Thu, 15 Aug 2019 20:42:59 +0800</pubDate><description><![CDATA[Problem Description
Given a sequence 1,2,3,…N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.
Input
Input contains multiple test cases. each case contai...]]></description><category></category></item><item><title><![CDATA[爬山下滑问题]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/99640620</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/99640620</guid><author>qq_26793545</author><pubDate>Thu, 15 Aug 2019 16:11:06 +0800</pubDate><description><![CDATA[Climbing Worm
Problem Description
An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. Durin...]]></description><category></category></item><item><title><![CDATA[递归]]></title><link>https://blog.youkuaiyun.com/qq_26793545/article/details/99611386</link><guid>https://blog.youkuaiyun.com/qq_26793545/article/details/99611386</guid><author>qq_26793545</author><pubDate>Wed, 14 Aug 2019 21:21:53 +0800</pubDate><description><![CDATA[ACM 规律递归 
Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input
...]]></description><category></category></item></channel></rss>