Ants 首先求出每个点的最近点. 可以直接对所有点构造kd树,然后在kd树上查询除本身以外的最近点,因为对所有点都求一次,所以不用担心退化. 也可以用分治做,同样是O(NlogN)的复杂度. 方法是每次用一条竖直线将平面划分成左右两半(平行线划分上下两半也可以),对两半中的点分别递归求每个点的最近点. 然后对左半边的一个点p,如果以它到左半边的最近点的距离为半径画出来的圆和右半边相交,则需要判断右半边是否有距离p更近的点. 我们求出这些圆和竖直平分线的交点,并将右半边的点投影到竖直平分线上,显…

2016 Al-Baath University Training Camp Contest-1 A题:http://codeforces.com/gym/101028/problem/A 题意:比赛初始值是1500,变化了几次,得到的正确结果和bug后的是否相等.(Tourist大佬好 Y(^o^)Y) #include <bits/stdc++.h> using namespace std; int main() { int t; cin>>t; while(t--) { in…

官方解题报告:http://bestcoder.hdu.edu.cn/blog/2015-multi-university-training-contest-6-solutions-by-zju/ 表示很难看....orz 1003题      链接:http://acm.hdu.edu.cn/showproblem.php?pid=5355 Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K…

1001. 一个数组上的两个区间求中位数,可以通过分类讨论直接找到中位数,复杂度O(1).不过本题数据较小,优美的log(n)也可过. 1002. 直接求得阴影面积表达式即可. 1003. 二分完成时间判是否可行.不妨设A的deadline比B的deadline短,对于一次判断内,由于总时间固定,两类任务完成的总时间固定,则可用来练习的时间也固定,首先找到可完成的收益最大的A类任务,若找不到则在B里找,此时若B类的收益还不如去练习,而且练习时间有剩余的话,就先去练习.A类任务若能做则不需要考虑收…

A Poor King Tag: Reversed BFS Preprocessing is needed to calculate answers for all positions (states). Beginning with all checkmate states, you should do reversed Breath-First Search (BFS) to update values. The state can be specified by three positio…

1001: 假设有4个红球,初始时从左到右标为1,2,3,4.那么肯定存在一种方案,使得最后结束时红球的顺序没有改变,也是1,2,3,4. 那么就可以把同色球都写成若干个不同色球了.所以现在共有n个颜色互异的球.按照最终情况标上1,2,..,n的序号,那么贪心的来每次操作就是把一个区间排序就行了. 1002: 环加树的同构计数问题.假如没有环,那么可以在树上dp来计算答案,具体方法是,把根节点的所有同构的子树看成相同的,然后用隔板法,算组合数即可得到答案.树的同构可以利用哈希来解决.有环的情况就…

A Boring Question \[\sum_{0\leq k_{1},k_{2},\cdots k_{m}\leq n}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}} \] \[=\sum_{0\leq k_{1}\leq k_{2}\leq\cdots \leq k_{m}\leq n}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}} \] \[=\sum_{k_{m}=0}^{n}\sum_{k_{m-1}=0}^{k…

ATM Mechine E(i,j):存款的范围是[0,i],还可以被警告j次的期望值. E(i,j) = \(max_{k=1}^{i}{\frac{i-k+1}{i+1} * E(i-k,j)+\frac{k}{i+1}*E(k-1,j-1)+1}\) 这样时间复杂度是\(O(K^2W)\)的. 假如Alice使用的是二分策略,那么在最坏情况下至多被警告\(\left \lceil log_{2}{K} \right \rceil\) 次. 于是W:=min(W,15)就可以了. 然后cla…

1001 Another Meaning 对于这个问题,显然可以进行DP: 令dp[i]表示到i结尾的字符串可以表示的不同含义数,那么考虑两种转移: 末尾不替换含义:dp[i - 1] 末尾替换含义:dp[i - |B|] (A.substr(i - |B| + 1,|B|) = B) 那么对于末尾替换含义的转移,需要快速判断\(B\)能不能和当前位置的后缀匹配,kmp或者hash判断即可. 复杂度:O(N) 1002 After a Sleepless Night 假设根已确定,可以发现新树若…

1001 Sqrt Bo 由于有\(5\)次的这个限制,所以尝试寻找分界点. 很容易发现是\(2^{32}\),所以我们先比较输入的数字是否比这个大,然后再暴力开根. 复杂度是\(O(\log\log n)\). 注意特判\(n=0\)的情况. 1002 Permutation Bo 根据期望的线性性,我们可以分开考虑每个位置对答案的贡献. 可以发现当\(i\)不在两边的时候和两端有六种大小关系,其中有两种是对答案有贡献的. 那么对答案的贡献就是\(\frac{c_i}{3}\). 在两端的话有…

Acperience 展开式子, \(\left\| W-\alpha B \right\|^2=\displaystyle\alpha^2\sum_{i=1}^{n}b_i^2-2\alpha\sum_{i=1}^n{w_ib_i}+\sum_{i=1}^{n}w_i^2\). 由于\(b_i\in \{+1,-1\}\), 那么\(\displaystyle\sum_{i=1}^{n}b_i^2=n\), 显然\(c=\displaystyle\sum_{i=1}^{n}w_i^2\)也是常…

首先向大家表示抱歉,因为这套题是去年出的,中间间隔时间太长,今年又临时准备仓促, 所以部分题目出现了一些问题,非常抱歉. Abandoned country 首先注意到任意两条边的边权是不一样的,由此得知最小生成树是唯一的,最小生成树既然 是唯一的,那么期望其实也就是唯一的,不存在什么最小期望.求完最小生成树之后,接下 来的问题就可以转换成在最小生成树上求任意两点之间距离的平均值,对于每条边,统计所 有的路径用到此边的次数,也就是边的两端的点数之积.那么这条边的总贡献就是次数*边 权.最后得到所…

Description ACM-SCPC-2017 is approaching every university is trying to do its best in order to be the Champion, there are n universities, the ith of them has exactly ai contestants. No one knows what is the official number of contestants in each team…

Description Tourist likes competitive programming and he has his own Codeforces account. He participated in lots of Codeforces Rounds, solved so many problems and became "Legendary Grand Master" (the highest rank on Codeforces). One day, he logg…

Description X is fighting beasts in the forest, in order to have a better chance to survive he's gonna buy upgrades for his weapon. Weapon upgrade shops are available along the forest, there are n shops, where the ith of them provides an upgrade with…

Description It is raining again! Youssef really forgot that there is a chance of rain in March, so he didn't fix the roof of his house. Youssef's roof is 1-D, and it contains n holes that make the water flow into the house, the position of hole i is…

 Description You've possibly heard about 'The Endless River'. However, if not, we are introducing it to you. The Endless River is a river in Cambridge on which David and Roger used to sail. This river has the shape of a circular ring, so if you kept…

Description The forces of evil are about to disappear since our hero is now on top on the tower of evil, and all what is left is the most evil, most dangerous monster! The tower has h floors (numbered from 1 to h, bottom to top), each floor has w roo…

Description Zaid has two words, a of length between 4 and 1000 and b of length 4 exactly. The word a is 'good' if it has a substring which is equal tob. However, a is 'almost good' if by inserting a single letter inside of it, it would become 'good'.…

Description X is well known artist, no one knows the secrete behind the beautiful paintings of X except his friend Y, well the reason that Y knows X's secrete is that he is that secret. Y is a programmer and he helps X with drawing paintings using co…

Description Rami went back from school and he had an easy homework about bitwise operations (and,or,..) The homework was like this : You have an equation : " A | B = C " ( this bar '|' means OR ) you are given the value of A and B , you need to…

Description A group of junior programmers are attending an advanced programming camp, where they learn very difficult algorithms and programming techniques! Near the center in which the camp is held, is a professional bakery which makes tasty pastrie…

比赛链接:http://codeforces.com/gym/101028/ 由于实习,几乎没有时间刷题了.今天下午得空,断断续续做了这一套题,挺简单的. A.读完题就能出结果. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */ #include <algorithm> #inclu…

题目链接:http://codeforces.com/problemset/gymProblem/101028/I I. March Rain time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output It is raining again! Youssef really forgot that there is a chance of…

Differencia Time Limit: 10000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 601    Accepted Submission(s): 173 Problem Description Professor Zhang has two sequences a1,a2,...,an and b1,b2,...,bn. He wants to…

Rigid Frameworks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 337    Accepted Submission(s): 273 Problem Description Erik Demaine is a Professor in Computer Science at the Massachusetts Insti…

8/11 2016 Multi-University Training Contest 1 官方题解 老年选手历险记 最小生成树+线性期望 A Abandoned country(BH) 题意: 1. 求最小生成树 2. 求在某一棵最小生成树任意两点的最小距离的期望值. 思路: 首先题目说了边权值都是不同的,所以最小生成树唯一.那么只要统计出最小生成树的每一条边在“任意两点走经过它“的情况下所贡献的值,发现在一棵树里,一条边所贡献的次数为,sz[v]表示v子树包括节点v的个数.如下图所示,红边所…

8/13 2016 Multi-University Training Contest 2官方题解 数学 A Acperience(CYD)题意: 给定一个向量,求他减去一个  α(>=0)乘以一个值为任意+1或-1的B向量后得到向量,求这个向量膜的最小值思路: 展开式子,当时最小,结果为. 代码: #include <bits/stdc++.h> using namespace std; long long w,a,b,c,a2; long long gcd(long long x,l…

5/11 2016 Multi-University Training Contest 3官方题解 2016年多校训练第三场 老年选手历险记 暴力 A Sqrt Bo(CYD) 题意:问进行多少次开根号向下取整能使给定的值为1,若为5次以上,则输出TAT: 思路: 有5次这个限制,得知值最大为10的11次方左右,存下后运算即可,注意初始值为0时运算的次数是大于5次的,输出TAT: 代码: #include <bits/stdc++.h> using namespace std; char st…

6/12 2016 Multi-University Training Contest 4官方题解 KMP+DP A Another Meaning(CYD) 题意: 给一段字符,同时给定你一个单词,这个单词有双重意思,字符串中可能会有很多这种单词,求这句话的意思总数:hehe. 思路: 可以用kmp算法快速求出串中的单词数量,若单词是分开的,每个单词有两种意思,可以直接相乘,若两个及以上单词在原串中是有交集的,那么数量不是直接相乘,发现这片连在一起的单词数量dp[i]=dp[i-1]+dp[j…