<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Paradise_tht的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Paradise_tht</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Paradise_tht]]></copyright><item><title><![CDATA[hdu1503 Advanced Fruits]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81637271</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81637271</guid><author>Paradise_tht</author><pubDate>Mon, 13 Aug 2018 21:17:30 +0800</pubDate><description><![CDATA[Problem Description 
The company “21st Century Fruits” has specialized in creating new sorts of fruits by transferring genes from one fruit into the genome of another one. Most times this method doesn...]]></description><category></category></item><item><title><![CDATA[ZOJ - 4006 Travel along the Line]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81637127</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81637127</guid><author>Paradise_tht</author><pubDate>Mon, 13 Aug 2018 21:08:56 +0800</pubDate><description><![CDATA[题意： 
从原点出发，求n秒后在m点的概率，每一秒有1/4可能向左移一格，1/4可能向右移一格，1/2可能不移动。 
T组数据，每一行输入n，m 
Sample Input 
3 
2 -2 
0 0 
0 1 
Sample Output 
562500004 
1 
0

用费马小定理求逆元。



#include&quot;bits/stdc++.h&quot;
using namespace std;
#d...]]></description><category></category></item><item><title><![CDATA[牛客网多校7 E	Counting 4-Cliques]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81589997</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81589997</guid><author>Paradise_tht</author><pubDate>Sat, 11 Aug 2018 19:23:13 +0800</pubDate><description><![CDATA[链接：https://www.nowcoder.com/acm/contest/145/E 
来源：牛客网

题目描述  
You love doing graph theory problems. You’ve recently stumbled upon a classical problem : Count the number of 4-cliques in an undirected g...]]></description><category></category></item><item><title><![CDATA[hdu6373 Pinball]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81513682</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81513682</guid><author>Paradise_tht</author><pubDate>Wed, 08 Aug 2018 18:55:49 +0800</pubDate><description><![CDATA[Problem Description 
There is a slope on the 2D plane. The lowest point of the slope is at the origin. There is a small ball falling down above the slope. Your task is to find how many times the ball ...]]></description><category></category></item><item><title><![CDATA[hdu1864 最大报销额]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81488714</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81488714</guid><author>Paradise_tht</author><pubDate>Tue, 07 Aug 2018 21:14:35 +0800</pubDate><description><![CDATA[Problem Description 
现有一笔经费可以报销一定额度的发票。允许报销的发票类型包括买图书（A类）、文具（B类）、差旅（C类），要求每张发票的总额不得超过1000元，每张发票上，单项物品的价值不得超过600元。现请你编写程序，在给出的一堆发票中找出可以报销的、不超过给定额度的最大报销额。

Input 
测试输入包含若干测试用例。每个测试用例的第1行包含两个正数 Q 和 N，其中 ...]]></description><category></category></item><item><title><![CDATA[hdu2955 Robberies]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81484348</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81484348</guid><author>Paradise_tht</author><pubDate>Tue, 07 Aug 2018 16:54:52 +0800</pubDate><description><![CDATA[题意 
有一个强盗要去几个银行偷盗，他既想多偷点钱，又想尽量不被抓到。已知各个银行的金钱数和被抓的概率，以及强盗能容忍的最大被抓概率。求他最多能偷到多少钱？ 
T组样例，p是强盗最大容忍被抓概率，n是有几家银行，接下n行，mi价值，pi是被抓概率 
0 &amp;lt; T &amp;lt;= 100，0.0 &amp;lt;= P &amp;lt;= 1.0，0 &amp;lt; N &amp;lt;= 100，0 &amp;lt; Mj &amp;lt;=...]]></description><category></category></item><item><title><![CDATA[2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 J.Beautiful Numbers]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81437445</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81437445</guid><author>Paradise_tht</author><pubDate>Mon, 06 Aug 2018 00:08:35 +0800</pubDate><description><![CDATA[题目： 
给你一个数，如果这个数能被这个数的各个数位上的数的和整除，那么这个数是beautiful number. 给你一个n, 求从1到n有多少个漂亮数。

输入： 
样例T 
每行一个n

输出： 
小于n的beautiful number 的个数(1 ≤ N ≤ 1e12).

一道数位dp题，感觉数位dp,一般sum值很大，数组肯定开不下，所以要尽可能的压缩sum的值，这道题假设n有len...]]></description><category></category></item><item><title><![CDATA[牛客网多校5 A	 gpa]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81393078</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81393078</guid><author>Paradise_tht</author><pubDate>Fri, 03 Aug 2018 19:17:55 +0800</pubDate><description><![CDATA[输入： 
第一行n,k 
第二行n个s[i] 
第三行n个c[i] 
最多剪掉k个数使最大

答案用二分找，做题时想到过，但没有仔细去想。假设max已知∑s[i]∗c[i]∑s[i]∗c[i]\sum s[i]*c[i]=max*∑s[i]∑s[i]\sum s[i],即∑s[i]∗(c[i]−max)∑s[i]∗(c[i]−max)\sum s[i]*(c[i]-max) = 0,只需要每次按(...]]></description><category></category></item><item><title><![CDATA[hdu6321 C. Dynamic Graph Matching]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81385383</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81385383</guid><author>Paradise_tht</author><pubDate>Fri, 03 Aug 2018 12:22:14 +0800</pubDate><description><![CDATA[问题描述： 
给定一个 n 个点的无向图，m 次加边或者删边操作。在每次操作后统计有多少个匹配包含 k = 1, 2, …,n/2 条边。 
2 ≤ n ≤ 10, 1 ≤ m ≤ 30000。

官方解释 


队友的代码，注释蛮详细的。。。



/** http://acm.hdu.edu.cn/showproblem.php?pid=6321
  * 题意：给n个点，求从中取没有公共点的几...]]></description><category></category></item><item><title><![CDATA[hdu6319 Problem A. Ascending Rating]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81366148</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81366148</guid><author>Paradise_tht</author><pubDate>Fri, 03 Aug 2018 12:05:28 +0800</pubDate><description><![CDATA[Problem A. Ascending Rating 
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) 
Total Submission(s): 4178    Accepted Submission(s): 1382

Problem Description 
Bef...]]></description><category></category></item><item><title><![CDATA[hdu6333 Problem B. Harvest of Apples]]></title><link>https://blog.youkuaiyun.com/Paradise_tht/article/details/81367008</link><guid>https://blog.youkuaiyun.com/Paradise_tht/article/details/81367008</guid><author>Paradise_tht</author><pubDate>Thu, 02 Aug 2018 20:42:46 +0800</pubDate><description><![CDATA[题意：一颗树上有n个苹果，求取不大于k个苹果的方案数。 
输入：T个案例，输入n,m，n,m&amp;amp;lt;=100000 
输出：方案数mod 1000000007

做法： 


方案数=∑mi=1C(i,n)∑i=1mC(i,n)\sum_{i=1}^{m} C(i,n),直接暴力会T，做的时候一直在想数学公式和找规律，没想到用莫队。n,m可以看成r,l，还有区间迁移的时候不是简单的加减，要找到区间...]]></description><category></category></item></channel></rss>