<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[DTL66的博客]]></title><description><![CDATA[欢迎大家评论探讨]]></description><link>https://blog.youkuaiyun.com/DTL66</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; DTL66]]></copyright><item><title><![CDATA[HDU 1394 Minimum Inversion Number【多数列逆序对】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53487813</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53487813</guid><author>DTL66</author><pubDate>Fri, 08 Mar 2019 13:44:01 +0800</pubDate><description><![CDATA[Minimum Inversion Number

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

Total Submission(s): 18906Accepted Submission(s): 11415





Problem Description...]]></description><category></category></item><item><title><![CDATA[HDU 3183 A Magic Lamp【RMQ区间取数（贪心）】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53573398</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53573398</guid><author>DTL66</author><pubDate>Sun, 11 Dec 2016 15:40:16 +0800</pubDate><description><![CDATA[A  Magic Lamp 
题意：给出一个大数，然后给出N,输出大数去除N个数之后的最小数，不输出前导0； 
思路：反向想一下就是从大数中取strlen(str)-N个数使取出的数最小，枚举每一个区间右端点，计算区间左端点即可，每次查询需要得到查找区间范围内最小值的位置（多个最小值时反回位置最小者），所以d[][]保存的应该是最小值的位置而不是最小值；*AC代码：#include<cstdio>]]></description><category></category></item><item><title><![CDATA[HDU 5443 The Water Problem【RMQ水题】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53564075</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53564075</guid><author>DTL66</author><pubDate>Sun, 11 Dec 2016 00:06:33 +0800</pubDate><description><![CDATA[The Water Problem

题意：给出N个数和Q次查询，没次查询L,R,输出【L,R】内数组元素的最大值；


AC代码：
#include
#include

using namespace std;

const int MAXN=1e3+11;
int a[MAXN],d[MAXN][32];

void RMQ_init(int N) {  
	for(int i=0]]></description><category></category></item><item><title><![CDATA[HDU 5289 Assignment【二分+RMQ】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53560352</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53560352</guid><author>DTL66</author><pubDate>Sat, 10 Dec 2016 19:14:22 +0800</pubDate><description><![CDATA[Assignment

题意：给一个数组，给一个K，找到满足连续子串的个数，使子串中的最小值与最大值之差小于K,  N：1e5
思路：确定右端点，二分枚举右端点，找到最大的满足位置，用RMQ判断，统计每一左端点的贡献即可；


AC代码：
#include
#include
#include

using namespace std;
typedef long long LL;

c]]></description><category></category></item><item><title><![CDATA[POJ 2823 Sliding Window【RMQ压缩（长度确定）】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53559797</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53559797</guid><author>DTL66</author><pubDate>Sat, 10 Dec 2016 17:24:34 +0800</pubDate><description><![CDATA[Sliding Window

题意：给一个长度为N的数组，输出所有区间长度为K的数组元素的最大值与最小值；


AC代码：
#include
#include

using namespace std;

const int MAXN=1e6+11;
int a[MAXN],dmax[MAXN],dmin[MAXN],ma[MAXN],mi[MAXN];

void RMQ_init]]></description><category></category></item><item><title><![CDATA[POJ 3264 Balanced Lineup【RMQ模板题】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53549010</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53549010</guid><author>DTL66</author><pubDate>Sat, 10 Dec 2016 00:30:38 +0800</pubDate><description><![CDATA[Assignment
题意：给出长度为N的数组和Q次查询，每次查询给出L,R，输出max(AL-AR)-min(AL-AR)； N：5e4,  Q：5e5;
模板题：需注意存储时如果从0开始那么查询时L-1,R-1;


AC代码：
#include
#include
#include

using namespace std;

const int MAXN=5e4+11;
int]]></description><category></category></item><item><title><![CDATA[LIGHT OJ 1138 - Trailing Zeroes (III)【N!后0的个数&&二分（好题）】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53538487</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53538487</guid><author>DTL66</author><pubDate>Fri, 09 Dec 2016 14:38:14 +0800</pubDate><description><![CDATA[1138 - Trailing Zeroes (III)





   
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Time Limit: 2 second(s)
Memory Limit: 32 MB





You task is to find minimal natural number N, so]]></description><category></category></item><item><title><![CDATA[CodeForces D. Nested Segments【逆序对类型】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53504206</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53504206</guid><author>DTL66</author><pubDate>Wed, 07 Dec 2016 12:42:01 +0800</pubDate><description><![CDATA[D. Nested Segments


time limit per test
2 seconds


memory limit per test
256 megabytes


input
standard input


output
standard output



You are given n segments on a line. Ther]]></description><category></category></item><item><title><![CDATA[LIGHT OJ 1245 - Harmonic Number (II) 【找规律】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53464572</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53464572</guid><author>DTL66</author><pubDate>Mon, 05 Dec 2016 13:47:07 +0800</pubDate><description><![CDATA[1245 - Harmonic Number (II)




   
PDF (English)
Statistics
Forum








Time Limit: 3 second(s)
Memory Limit: 32 MB





I was trying to solve problem '1234 - Harmonic Numbe]]></description><category></category></item><item><title><![CDATA[HDU 1166-敌兵布阵【树状数组&&线段树单点更新】【模板】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53454788</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53454788</guid><author>DTL66</author><pubDate>Sun, 04 Dec 2016 16:18:13 +0800</pubDate><description><![CDATA[敌兵布阵

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

Total Submission(s): 80748    Accepted Submission(s): 34104




Problem Description

C国的死对头A国这段时间正]]></description><category></category></item><item><title><![CDATA[NYOJ 977 最大的最小公倍数]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53454374</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53454374</guid><author>DTL66</author><pubDate>Sun, 04 Dec 2016 15:21:36 +0800</pubDate><description><![CDATA[最大的最小公倍数
时间限制：1000 ms  |  内存限制：32768 KB
难度：2





描述
　　高中时我们对最小公倍数就已经很熟悉了，相信你很快就可以把这个问题解决。这次的问题是：给你一个正整数n，任取三个不大于n的正整数，取法不限，每个数可取多次，使得取到的这三个数的最小公倍数在所有取法中是最大的。
　　例如当n = 5 时，不大于5的数为1、2、3、4、5。则]]></description><category></category></item><item><title><![CDATA[HPU 1195: Mod [STL、二分、单调队列、DFS]]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53440875</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53440875</guid><author>DTL66</author><pubDate>Sat, 03 Dec 2016 00:58:56 +0800</pubDate><description><![CDATA[Mod
时间限制：4000 ms  |  内存限制：65535 KB
难度：2





描述
mod是取余运算，在程序中用符号"%"来表示。


如3 % 7 = 3，7 % 5 = 2，0 % 4 = 0。


Ocean用巧妙的方法得到了一个序列，该序列有NN 个元素，我们用数组aa 来记录（下标从00 到N−1


Ocean定义f[i] = (((i ]]></description><category></category></item><item><title><![CDATA[HPU 1200: 我说你猜]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53434257</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53434257</guid><author>DTL66</author><pubDate>Fri, 02 Dec 2016 14:23:15 +0800</pubDate><description><![CDATA[1200: 我说你猜 [水题]
时间限制: 1 Sec  内存限制: 128 MB
提交: 285  解决: 44
[提交][状态][讨论版]

题目描述

“三下二除五”，答案是一个浮点数。

输入

没有输入

输出

请输出最吻合的答案，不能有任何多余的输出。
比如要输出2.5，你不能在末尾输出多余的0。输出2.50或者2.500会被认为是错误的。


]]></description><category></category></item><item><title><![CDATA[HPU 1029: QAQ的填充方案 【Catlan】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53426337</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53426337</guid><author>DTL66</author><pubDate>Thu, 01 Dec 2016 20:26:30 +0800</pubDate><description><![CDATA[1029: QAQ的填充方案 [组合数学]
时间限制: 1 Sec  内存限制: 128 MB
提交: 19  解决: 6
[提交][状态][讨论版]

题目描述

给定1−2∗N1−2∗N共2∗N2∗N个元素，现在有x[1−N]，y[1−N]x[1−N]，y[1−N]两个数组，你需要把这2∗N2∗N个数（每个数必须用一次且仅用一次）填入这两个数组里面。
定义一个填充方案是合法的即]]></description><category></category></item><item><title><![CDATA[HPU 1046: QAQ的数学问题 【贝祖定理】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53420968</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53420968</guid><author>DTL66</author><pubDate>Thu, 01 Dec 2016 14:25:28 +0800</pubDate><description><![CDATA[1046: QAQ的数学问题 [数学]
时间限制: 1 Sec  内存限制: 128 MB
提交: 61  解决: 11
[提交][状态][讨论版]

题目描述

QAQ很喜欢数学，尤其对LCMLCM（最小公倍数）很感兴趣。
对于数对(6,10)(6,10)，可以得出LCM(6,10)=30LCM(6,10)=30。为了让LCMLCM的值最小化，他尝试把66和1010全部加上22，]]></description><category></category></item><item><title><![CDATA[UVA - 10943 How do you add?【组合数取余（递推）】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53285862</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53285862</guid><author>DTL66</author><pubDate>Tue, 22 Nov 2016 14:52:05 +0800</pubDate><description><![CDATA[Larry is very bad at math — he usually uses a calculator, whichworked well throughout college. Unforunately, he is now struck ina deserted island with his good buddy Ryan after a snowboardingaccident.]]></description><category></category></item><item><title><![CDATA[UVA - 11388 GCD LCM]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53284419</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53284419</guid><author>DTL66</author><pubDate>Tue, 22 Nov 2016 12:33:44 +0800</pubDate><description><![CDATA[The GCD of two positive integers is the largest integer that divides both the integers without any
remainder. The LCM of two positive integers is the smallest positive integer that is divisible by bo]]></description><category></category></item><item><title><![CDATA[UVA - 11889 Benefit【LCM】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53284388</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53284388</guid><author>DTL66</author><pubDate>Tue, 22 Nov 2016 12:28:06 +0800</pubDate><description><![CDATA[Recently Yaghoub is playing a new trick to sell some more. When somebody gives him A Tomans, he
who never has appropriate changes, asks for B Tomans such that lowest common multiple of A and B
equal]]></description><category></category></item><item><title><![CDATA[Codeforce 729B. Spotlights]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53265850</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53265850</guid><author>DTL66</author><pubDate>Mon, 21 Nov 2016 22:28:09 +0800</pubDate><description><![CDATA[B. Spotlights


time limit per test
1 second


memory limit per test
256 megabytes


input
standard input


output
standard output



Theater stage is a rectangular field of size n]]></description><category></category></item><item><title><![CDATA[POJ 3278 Catch That Cow【BFS】]]></title><link>https://blog.youkuaiyun.com/DTL66/article/details/53170139</link><guid>https://blog.youkuaiyun.com/DTL66/article/details/53170139</guid><author>DTL66</author><pubDate>Tue, 15 Nov 2016 14:02:07 +0800</pubDate><description><![CDATA[Catch That Cow




Time Limit: 2000MS
 
Memory Limit: 65536K


Total Submissions: 79281
 
Accepted: 25001





Description

Farmer John has been informed of the location of a fugit]]></description><category></category></item></channel></rss>