<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[yulubupt的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/yulubupt</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; yulubupt]]></copyright><item><title><![CDATA[POJ1007解题报告]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/39142661</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/39142661</guid><author>yulubupt</author><pubDate>Tue, 09 Sep 2014 00:02:58 +0800</pubDate><description><![CDATA[#include 
#include 
using namespace std;

typedef struct DNA
{
	string s;
	int value;
}DNA;

int cmp(const DNA *a,const DNA *b)
{
	return (a->value-b->value);
}

int main()
{
	int n,m;
	while(cin>>n>>]]></description><category></category></item><item><title><![CDATA[算法导论 chapter 10.1  Page233 Stack push and pop]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/37935393</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/37935393</guid><author>yulubupt</author><pubDate>Fri, 18 Jul 2014 18:16:39 +0800</pubDate><description><![CDATA[//***算法导论 chapter 10 Elementary Data Structures***///
//Page 233 伪代码转换成代码
//Stack push and pop

#include 
#define TRUE 1
#define FALSE 0
using namespace std;

int stack_array[20]={0};//定义一个数组，当做桟来使用 
]]></description><category></category></item><item><title><![CDATA[堆排序 Heapsort]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25991593</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25991593</guid><author>yulubupt</author><pubDate>Fri, 16 May 2014 16:50:39 +0800</pubDate><description><![CDATA[/*=============================================================================
#
#     FileName: HeapSort.cpp
#         Desc: 堆排序。数组的序号从1到len，长度为len+1。 
#
#       Author: yulu
#        Email: 1873737]]></description><category></category></item><item><title><![CDATA[寻找最大子数组   采用线性的方法]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25336965</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25336965</guid><author>yulubupt</author><pubDate>Thu, 08 May 2014 21:30:06 +0800</pubDate><description><![CDATA[题目来源： 《算法导论》]]></description><category></category></item><item><title><![CDATA[华为OJ  在字符串中找出连续最长的数字串]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25313069</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25313069</guid><author>yulubupt</author><pubDate>Thu, 08 May 2014 15:52:41 +0800</pubDate><description><![CDATA[第一种算法 O（n*n），时间超时了。不过答案是对的。
#include 
#include 
#include 
//#include "oj.h"
using namespace std;


/* 功能：在字符串中找出连续最长的数字串，并把这个串的长度返回
函数原型：
   unsigned int Continumax(char** pOutputstr,  char* intputst]]></description><category></category></item><item><title><![CDATA[寻找最大和子数组 Find maximum subarrray]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25302055</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25302055</guid><author>yulubupt</author><pubDate>Thu, 08 May 2014 14:09:15 +0800</pubDate><description><![CDATA[问题描述：输入是一个大小为n的整型数组，要求输出数组的任何连续子数组中的最大值。例如：输入的数组为array[10]
 = {31,-41,59,26,-53,58,97,-93,-23,84};输出最大连续子数组和为array[2...6]：187]]></description><category></category></item><item><title><![CDATA[华为OJ  Home+work]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25081823</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25081823</guid><author>yulubupt</author><pubDate>Mon, 05 May 2014 22:58:17 +0800</pubDate><description><![CDATA[#include "OJ.h"
 
  
/*

输入: nPapers表示试卷的数目(1≤Papers≤20)，nRemain表示剩余的时间(1≤nRemain≤10000)，paper[][2]是一个Papers*2的数组，

每一行的两个元素依次为做完这一份试卷所需的时间、做完这份试卷的价值

输出: *pMaxValue为获得的最大价值
返回:
0:异常
1:计算成功返回


*/
int]]></description><category></category></item><item><title><![CDATA[华为OJ  向升序单向链表中插入一个节点]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/25011595</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/25011595</guid><author>yulubupt</author><pubDate>Mon, 05 May 2014 00:59:02 +0800</pubDate><description><![CDATA[直接替代该函数的代码即可。无需tian'j'ai
ListNode* InsertNodeToList(ListNode* pListHead, ListNode* pInsertNode)
{
	if(!pListHead || !pInsertNode)
	    return (ListNode *)NULL;
    
    ListNode *p=pListHead;
    //在]]></description><category></category></item><item><title><![CDATA[冒泡排序 Bubble Sort]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/24852725</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/24852725</guid><author>yulubupt</author><pubDate>Thu, 01 May 2014 23:51:42 +0800</pubDate><description><![CDATA[效率比较低，时间复杂度应该是O(]]></description><category></category></item><item><title><![CDATA[结合归并排序和插入排序 Merge with Insertion Sort]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/24852217</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/24852217</guid><author>yulubupt</author><pubDate>Thu, 01 May 2014 23:31:21 +0800</pubDate><description><![CDATA[对n(2的整数幂个数字进行排序，]]></description><category></category></item><item><title><![CDATA[归并排序 Merge sort]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/24819905</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/24819905</guid><author>yulubupt</author><pubDate>Thu, 01 May 2014 00:29:24 +0800</pubDate><description><![CDATA[时间复杂度 nlogn]]></description><category></category></item><item><title><![CDATA[线性表的链式存储]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/23183529</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/23183529</guid><author>yulubupt</author><pubDate>Tue, 08 Apr 2014 14:43:55 +0800</pubDate><description><![CDATA[/******线性表链式存储******/
/*来源：《大话数据结构》*/
#define OK 1
#define ERROR 1
typedef int Status;

typedef int ElemType; 
/*线性表的单链表存储结构*/ 
typedef struct Node
{
	ElemType data;
	struct Node *next;
}Node;

typede]]></description><category></category></item><item><title><![CDATA[线性表的顺序存储]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/23166487</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/23166487</guid><author>yulubupt</author><pubDate>Tue, 08 Apr 2014 10:16:32 +0800</pubDate><description><![CDATA[/******线性表的顺序存储******/
/*来源：《大话数据结构》*/
 
#include 
#define MAXSIZE 10 /*存储空间初始分配*/
using namespace std;

#define OK 1
#define ERROR 0
#define TRUE 1
#define FALSE 0
typedef int Status;

//线性表的存储结构 
ty]]></description><category></category></item><item><title><![CDATA[POJ2255]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/20064175</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/20064175</guid><author>yulubupt</author><pubDate>Thu, 27 Feb 2014 18:27:47 +0800</pubDate><description><![CDATA[……


#include
#include
#include

typedef struct Node{
	char ch;
	struct Node *left;
	struct Node *right;
}Node; 

char mid[27];
char pre[27];
int n = -1;

void make_tree(int i, int j);

int main(vo]]></description><category></category></item><item><title><![CDATA[POJ3006 解题报告]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/19637623</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/19637623</guid><author>yulubupt</author><pubDate>Fri, 21 Feb 2014 20:09:00 +0800</pubDate><description><![CDATA[#include
#include
using namespace std;

const int MAX=1000000;
bool prime[MAX];
int main()
{
	memset(prime,1,MAX);
	prime[0]=0;
	prime[1]=0;
	for(int i=2;i<MAX;i++)
	{
		if(prime[i])
		{
			for(int j=]]></description><category></category></item><item><title><![CDATA[筛法求素数]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/19637227</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/19637227</guid><author>yulubupt</author><pubDate>Fri, 21 Feb 2014 19:58:30 +0800</pubDate><description><![CDATA[#include
#include
using namespace std;

const int MAX=1000000;
bool prime[MAX];
int main()
{
	memset(prime,1,MAX);
	prime[0]=0;
	prime[1]=0;
	for(int i=2;i<MAX;i++)
	{
		if(prime[i])
		{
			for(int j=]]></description><category></category></item><item><title><![CDATA[POJ1503 解题报告]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/19634887</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/19634887</guid><author>yulubupt</author><pubDate>Fri, 21 Feb 2014 18:51:55 +0800</pubDate><description><![CDATA[解题思路：
首先将字符串存在char a[]当中，然后将数反转过来，并且转化为int型，存在temp[]中。
每次用新的temp与result做加法运算，该进位的要进位。
需要注意的是，不能把数组开得太小，以下代码在temp[110], result[110]时就Wrong Answer了，但是改成150就AC了。


#include
#include
#include
using ]]></description><category></category></item><item><title><![CDATA[POJ1083 解题报告]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/19426777</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/19426777</guid><author>yulubupt</author><pubDate>Tue, 18 Feb 2014 21:05:05 +0800</pubDate><description><![CDATA[解题思路：
如果一个门前同时有两个桌子经过，则该操作不能同时进行。例如1--4与3--6，则4的门前有两个桌子同时经过，3的门前也同时有两个桌子经过，则该操作不能同时进行。
注意，3与4是同一个地方。




#include
using namespace std;

int main()
{
	int TestCases;//The number of test cases.
	]]></description><category></category></item><item><title><![CDATA[POJ2739 解题报告]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/19087223</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/19087223</guid><author>yulubupt</author><pubDate>Tue, 11 Feb 2014 20:46:28 +0800</pubDate><description><![CDATA[解题思路:
1. 构造一个素数表
    求1-10000里面的素数，存储在prime[10000]数组当中。我们用boo[i]来记录——如果i为素数，则boo[i]=0；反之，则boo[i]=1。
    对于1-10000之间的任意一个数i，如果i能被2~（i的开方）之中的任何一个数字整除，则i不是素数；反之，则i为素数。因此我们用j的循环来判断i是不是素数。
    最后，只要判断b]]></description><category></category></item><item><title><![CDATA[494]]></title><link>https://blog.youkuaiyun.com/yulubupt/article/details/16945219</link><guid>https://blog.youkuaiyun.com/yulubupt/article/details/16945219</guid><author>yulubupt</author><pubDate>Mon, 25 Nov 2013 15:34:15 +0800</pubDate><description><![CDATA[#includeint main(){char str[1000];char *s;int n=0,flag=1;while( gets(str)!=NULL ){s=str;while( *s!='\0' ){if( (*s>='A' && *s='a' && *s<='z') ){if( flag==1 ){ n++; flag=0;}}else flag=1;}printf("%d\n",n]]></description><category></category></item></channel></rss>