题目:http://acm.hdu.edu.cn/showproblem.php?pid=5594 完全不会啊TAT.. 其实官方题解已经说的很清楚了.. #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cstdlib> #include <queue> #include <cmath> #def…

ZYB's Prime Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5594 Description After getting 600 scores in NOIP,ZYB(ZJ−267) creates a problem:you are given N numbers,now you are asked to divide them into K groups(…

题目简述:有一个全排列,一直每个前缀区间的逆序对数,还原这个排列. fi记录逆序对数,pi记录该位置数值,则k=fi-f(i-1)表示前i-1个数比pi大的数的个数,那么只要在剩余元素求出按大小顺序第i-k个数字即可. 线段树+二分搜索,线段树bit[i]记录i的在剩余元素的排名顺序. /******************************* Date : 2015-12-06 19:49:59 Author : WQJ (1225234825@qq.com) Link : http:/…

ZYB's Premutation Accepts: 220 Submissions: 983 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) 问题描述 ZYBZYBZYB有一个排列PPP,但他只记得PPP中每个前缀区间的逆序对数,现在他要求你还原这个排列. (i,j)(i<j)(i,j)(i < j)(i,j)(i<j)被称为一对逆序对当且仅当Ai>AjA_i&g…

ZYB's Game Accepts: 672 Submissions: 1207 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description ZYBZYBZYB played a game named NumberBombNumber BombNumberBomb with his classmates in hiking:a host keeps a…

ZYB's Biology Accepts: 848 Submissions: 1199 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description After getting 600600600 scores in NOIPNOIPNOIP ZYB(ZJ−267)ZYB(ZJ-267)ZYB(ZJ−267) begins to work with b…

ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 638    Accepted Submission(s): 302 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each pr…

主持人一直某个数字在1到n范围,假设甲乙已经知道,甲先猜乙后,都采用最优策略,主持人说偏大还是偏小,不断缩小范围,问最后乙能会获胜的X的取值的个数. 如果n为奇数,那么仅当x=n/2乙必然获胜,若为奇数乙不可能获胜. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue…

本文介绍CVE-2022 0778漏洞及其复现方法,并精心构造了具有一个非法椭圆曲线参数的证书可以触发该漏洞. 本博客已迁移至CatBro's Blog,那是我自己搭建的个人博客,欢迎关注.本文链接 漏洞描述[1] 漏洞出自BN_mod_sqrt()接口函数,它用于计算模平方根,且期望参数p应该是个质数,但是函数内并没有进行检查,这导致内部可能出现无限循环.这个函数在解析如下格式的证书时会被用到: 证书包含压缩格式的椭圆曲线公钥时 证书带有显式椭圆曲线参数,其基点是压缩格式编码的 总之,在解析证…

String[] img = "FF D8 FF FE 00 24 47 00 9D 0C 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 F0 00 40 01 48 00 32 12 0B 51 04 51 04 00 00 FF DB 00 84 00 12 0C 0E 10 0E 0B 12 10 0F 10 14 13 12 15 1B 2D 1D 1B 19 19 1B 37 27 2A 21 2D 41 39 45 44 40 39 3F…

传送门 直接求连通的不好做,考虑容斥 设 \(g_i\) 表示至少有 \(i\) 个连通块的方案数,\(f_i\) 表示恰好有 \(i\) 个的 那么 \[g_x=\sum_{i=x}^{n}\begin{Bmatrix}x \\ i\end{Bmatrix}f_i\iff f_x=\sum_{i=x}^{n}(-1)^{i-x}\begin{bmatrix}x \\ i\end{bmatrix}g_i\] 那么 \[f_1=\sum_{i=1}^{n}(-1)^{i-1}(i-1)!g_i\]…

问题背景 业务准备在天翼云上搭建一套线上环境,VM 操作系统版本为 CentOS Linux release 7.4.1708,但是在 ambari Web 管理页面上部署hadoop节点主机的时候,遇到了register失败,无法继续部署的问题. 安装 Ambari 的程序包都是在老的集群环境直接拷贝过来,在老集群上(Red Hat Enterprise Linux Server release 7.3 (Maipo))并未出现此问题,OpenSSL version OpenSSL 1.0.2…

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 354    Accepted Submission(s): 100 Problem Description ZYB has a tree with N nodes,now he wants you to solve the numbers of nodes distanced no m…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 175    Accepted Submission(s): 74 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutat…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5593 点分治被卡了TAT... 正解是dp,可以按层数考虑dp,先预处理跑一边dfs得到子树各层数点数大小,再dfs一遍找出不在子树上的与当前点距离<=k的点. 设u为当前点,v为其一个儿子,有g[v][1]=1 g[v][i]+=g[u][i-1](i>=2) g[v][i]+=f[u][i-1]-f[v][i-2](i>=2) 完全不难嘛比赛的时候人都傻了TAT,还是太弱了 #include…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5592 题目大意就是给了每个[1, i]区间逆序对的个数,要求复原原序列. 比赛的时候2B了一发. 首先既然给了[1, i-1]和[1, i]区间逆序对的个数,自然可以求出与i组成逆序对的个数,自然就是i前面比i大的个数,自然也就能求出i前面比i小的个数. 然后我考虑最后一个数,由于它在最后一个,所以所有数都在它前面,自然如果知道它前面有多少个比他小的,就知道它是几了.不妨设最后一个是p,然后考虑倒数…

Vega Prime 1.2 (视景仿真) MPI的视景仿真渲染工具Vega是世界上领先的应用于实时视景仿真.声音仿真和虚拟现实等领域的软件环境,它用来渲染战场仿真.娱乐.城市仿真.训练模拟器和计算可视化等领域的视景数据库,实现环境效果等的加入和交互控制.它将易用的工具和高级视景仿真功能巧妙地结合起来,从而可使用户简单迅速地创建.编辑.运行复杂的实时三维仿真应用.由于它大幅度减少了源代码的编写,使软件的进一部维护和实时性能的优化变得更容易,从而大大提高了开发效率.使用它可以迅速地创建各种实时交互…

感觉其实就是树分治,一次BC的题,感觉这次题目质量比较高,仅代表蒟蒻的看法 一次DFS获取每个点到子树的距离不大于K的点的个数, 然后一遍BFS获取从每个点父亲不大于K的的个数,层层扩展,还是想说 其实就是树分治.....并没有什么DP /* Problem : 5593 ( ZYB's Tree ) Judge Status : Accepted RunId : 15764784 Language : G++ Author : qianbi08 */ #include<cstdio> #in…

Building a Space Station Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5699   Accepted: 2855 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You ar…

Prime Ring Problem Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 18   Accepted Submission(s) : 7 Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2,…

Problem Description ZYB played a game named NumberBomb with his classmates ,N] in mind,then players guess a number in turns,the player who exactly guesses X loses,or the host will tell all the players that the number now ,N] at first,each player shou…

Problem Description After getting scores ) begins to work with biological questions.Now he give you a simple biological questions: he gives you a DNA sequence and a RNA sequence,then he asks you whether the DNA sequence and the RNA sequence are match…

ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 758    Accepted Submission(s): 359 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each pr…

几年的股市可谓惨不忍睹,不提也罢.唯有打新中签的时候,心里稍微有那么一点点的补偿,于是内心就YY可以30板吗,可以40板吗.于是就写了个连板的bc程序,每次中签的时候就运行一下,然后尽情的YY,然而每次还是无法利益最大化,尽管也就中了这么两次. 喜欢bc这个计算器,因为bc近似于C的语法和任意精度.以下程序算新股连板股价绝对精确,供君YY. #!/usr/bin/bc scale = 2 x = read(); times = read(); y = x*100*144; scale = 0;…

bc是Linux下的命令行式的计算器. 题目虽然叫任意进制,但是因为bc的限制,输入进制是2~16范围:输出进制是2~999范围.这与常见计算器的进制范围是一致的,比如windows计算器最高也只能处理16进制输入数据. 一.bc计算器的使用 bc计算器默认输入.输出都为10进制. [root@centos6 ~]# bc #打开bc计算器 bc 1.06.95 Copyright 2006 Free Software Foundation, Inc. This is free software…

适合对并查集有一定理解的人.  新手可能看不懂吧.... 并查集简单点说就是将相关的2个数字联系起来 比如 房子                      1   2    3   4  5   6 能通向的房子        2   3    4  5  6    1 主要 建立并查集的 函数结构 模板(一般不变除非加权--最好能理解) for(int i=0;i<n;i++)         flag[i]=i;               //标记数组初始化以方便寻根 1 int find…

1.计算(a/b)%c,其中b能整除a 设a=b*r=(bc)*s+b*t 则(b*t)为a除以bc的余数 r=c*s+t 而 (a/b)%c=r%c=t (a%bc)/b=(b*t)/b=t 所以对于b与c互素和不互素都有(a/b)%c=(a%bc)/b成立. 当bc不大时,先取模bc,再除b 如果b与c互素,则(a/b)%c=a*b^(phi(c)-1)%c 待证 2.与集合子集 斐波那契数列的第n+2项同时也代表了集合{1,2,...,n}中所有不包含相邻正整数的子集个数. 证明:归纳法证…

ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 637    Accepted Submission(s): 301 Problem Description ZYB has a premutation P,but he only remeber the reverse log of each pre…

ZYB's Tree Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 400    Accepted Submission(s): 114 Problem Description ZYB has a tree with N nodes,now he wants you to solve the numbers of nodes dis…

Building a Space Station Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6635   Accepted: 3236 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You ar…