题意:有一个集合,求有多少形态不同的二叉树满足每个点的权值都属于这个集合并且总点权等于i 题解:先用生成函数搞出来\(f(x)=f(x)^2*c(x)+1\) 然后转化一下变成\(f(x)=\frac{2}{1+\sqrt{1-4*c(x)}}\) 然后多项式开根和多项式求逆即可(先对下面的项开根,然后再求逆) 多项式开根: \(B(x)^2=A(x) \bmod x^{ \lfloor \frac{n}{2} \rfloor}\) \(B'(x)^2=A(x) \bmod x^{ \lfloo…

D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/D Description At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important…

B. The Child and Zoo Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/B Description Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animal…

A. The Child and Toy Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/A Description On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy…

题目链接:http://codeforces.com/problemset/problem/438/D 给你n个数,m个操作,1操作是查询l到r之间的和,2操作是将l到r之间大于等于x的数xor于x,3操作是将下标为k的数变为x. 注意成段更新的时候,遇到一个区间的最大值还小于x的话就停止更新. #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __…

     好题啊,被HACK了.曾经做题都是人数越来越多.这次比赛 PASS人数 从2000直掉 1000人  被HACK  1000多人! ! ! ! 没见过的科技啊 1 2 4 8 这组数 被黑的 A. The Child and Homework time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Once upon a ti…

D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of…

D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of…

D. The Child and Zoo time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains …

B. The Child and Set   At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite set of Picks. Fortunately, Picks remembers something about…

D. The Child and Sequence   At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks. Fortunately, Picks remembers how t…

题目例如以下: C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't…

注意题目长度不能考虑前缀,而且如果即存在一个选项的长度的两倍小于其他所有选项的长度,也存在一个选项的长度大于其他选项长度的两倍,则答案不是一个好的选择,只能选择C. #include <iostream> #include <vector> #include <string> #include <algorithm> using namespace std; struct Answer{ char item; int length; Answer():ite…

<传送门> [题目大意] 给你一个sum和一个limit,现在要你在1~limit中找到一些数来使得这些数的和等于sum,如果能找到的话就输出找到的数的个数和这些数,未找到输出"-1". 比赛的时候被hack了. [题目分析] 这题需要将所有的数的lowbit先求出来,然后按照大小排序,然后从后往前判断,如果这个数小于sum那么这个数就是可以构成sum的数,选进去,完了以后判断sum的值是否为0就可以了.做题的时候没将题目理解透彻. #include<bits/std…

注意此题,每一个部分都有一个能量值v[i],他移除第i部分所需的能量是v[f[1]]+v[f[2]]+...+v[f[k]],其中f[1],f[2],...,f[k]是与i直接相连(且还未被移除)的部分的编号. 注意题目移除的都是与第i部分直接相连的部分的能量值, 将本题目简化得,只考虑两个点1和2,1和2相连,1的能量值是10,2的能量值是20, 移除绳子时,要保持能量最小,可以移除部分2,这样移除的能量就是与2相连的部分1的能量即是10: 故每次相连两部分都移除能量值大的即可 #includ…

题目链接 题意: 给定goal和limit,求1-limit中的若干个数,每一个数最多出现一次,且这些数的lowbit()值之和等于goal,假设存在这种一些数,输出个数和每一个数:否则-1 分析: 先考虑一下比較普通的情况,给一些数,和一个goal,问时候能达到.(最好还是设这些数已经从大到小排序) 考虑能否够贪心,对于当前的数x: 1.之后的数的和能等于x,那么假设x<=goal,显然必须选x: 2.之后的数的和能等于x-1,那么同上(这个情况就是二进制的情况) 3.之后的数的和不包含上述两…

链接:http://codeforces.com/contest/437/problem/A A. The Child and Homework time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Once upon a time a child got a test consisting of multiple-choice qu…

感觉不会再爱了,呜呜! A题原来HACK这么多!很多人跟我一样掉坑了! If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. If there is exactly one…

A. The Child and Homework time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question…

给你一张无向图,每个点有一个权值,对于一条从l到r 的边权值是l到r路径上最小的点的权值,(多条路取最大的权值),然后求每两个点之间的权值和/点对数 题解:并查集维护,先从点大的边排序,然后依次加边,这样每次加进来的保证是当前最大 的,然后每次合并都要加上两端的最小值*两端的size,类似与每一个最小值算贡献 #pragma comment(linker, "/stack:200000000") #pragma GCC optimize("Ofast,no-stack-pro…

这几次CF都挺惨.. A 没条边权设为两端点的最小点权,最后加起来. 数组开小,WA一次 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set>…

题目大意:给出一个森林,每次询问给出u,v,问从u所在连通块中随机选出一个点与v所在连通块中随机选出一个点相连,连出的树的直径期望(不是树输出-1).(n,q<=10^5) 解法:预处理出各连通块的直径和各点到连通块内一点的最远距离d[x](树形dp+换根),询问若在同一块内输出-1,否则若随机选出两点x,y,直径为max(d[x]+d[y]+1,x所在块直径,y所在块直径),我们把同一连通块内的d排序,枚举小的连通块中的d,到大的连通块中二分d[x]+d[y]+1<=max(x所在块直径,y…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 把n分解成二进制的形式. n=2^a0+2^a1+...+2^a[q-1] 则固定就是长度为q的序列. 要想扩展为长为k的序列. 可以把2^x转化为2^(x-1)+2^(x-1)的形式. 这样序列的长度就+1了 它要求max{ai}最小 那么我们可以枚举ai的最大值是什么->i (递减着枚举) 然后比i大的ai都换成两个ai-1的形式. 然后看看序列的长度是否小于等于k; 如果小于k的话. 就把min{ai}分解成两个min{a…

[链接]点击打开链接 [题意] 给你一棵n个点的树,每个点的美丽值定义为根节点到这个点的路径上的所有权值的gcd. 现在,假设对于每一个点,在计算美丽值的时候,你可以将某一个点的权值置为0的话. 问你每个点的最大美丽值可能是多少. [题解] 从根节点开始进行dfs,在往下走的过程中,暴力用set记录下路径中把以上的每一个点删掉后的gcd是什么. 这set里的东西.进行扩展的时候,只能加入不删的点.因为它已经表示删掉一个点的状态了.下面只能都不删 了.然后用另外一个数组now,记录下到根节点到当前…

Problem  Codeforces Round #556 (Div. 2) - D. Three Religions Time Limit: 3000 mSec Problem Description Input Output Sample Input 6 8abdabc+ 1 a+ 1 d+ 2 b+ 2 c+ 3 a+ 3 b+ 1 c- 2 Sample Output YESYESYESYESYESYESNOYES 题解:动态规划,意识到这个题是动态规划之后难点在于要优化什么东西,本题…

Codeforces Round #418 (Div. 2) D. An overnight dance in discotheque 题意: 给\(n(n <= 1000)\)个圆,圆与圆之间不存在相交关系,只存在包含与不包含关系,现在问你把这一堆圆分成两组,求每组圆的异或并和的最大值. 思路:最简单的做法是贪心的做法,算出每个圆被包含的次数,为偶数则贡献为正,奇数为贡献为负. 现在主要学习一下树形dp的做法 由于圆不存在相交的情况,所以可以把所有的圆建成树,即把每棵树上的点分成两组,分别建成…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…