#1391 : Countries 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 There are two antagonistic countries, country A and country B. They are in a war, and keep launching missiles towards each other. It is known that country A will launch N missiles. The i-th miss…

#1392 : War Chess 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 Rainbow loves to play kinds of War Chess games. There are many famous War Chess games such as "Biography of Cao Cao", "Anecdotes of Wagang Mountain", etc. In this problem, let's c…

#1389 : Sewage Treatment 时间限制:2000ms 单点时限:2000ms 内存限制:256MB 描述 After years of suffering, people could not tolerate the pollution in the city any more, and started a long-lasting protest. Eventually, the municipal government made up its mind to deal w…

#1586 : Minimum 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 You are given a list of integers a0, a1, …, a2^k-1. You need to support two types of queries: 1. Output Minx,y∈[l,r] {ax∙ay}. 2. Let ax=y. 输入 The first line is an integer T, indicating the number…

#1582 : Territorial Dispute 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In 2333, the C++ Empire and the Java Republic become the most powerful country in the world. They compete with each other in the colonizing the Mars. There are n colonies on the Mars,…

#1584 : Bounce 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 For Argo, it is very interesting watching a circle bouncing in a rectangle. As shown in the figure below, the rectangle is divided into N×M grids, and the circle fits exactly one grid. The bouncing…

#1578 : Visiting Peking University 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 Ming is going to travel for n days and the date of these days can be represented by n integers: 0, 1, 2, …, n-1. He plans to spend m consecutive days(2 ≤ m ≤ n)in Beijing. Durin…

https://hihocoder.com/problemset/problem/1586 线段树操作,原来题并不难..... 当时忽略了一个重要问题,就是ax*ay要最小时,x.y可以相等,那就简单了!结果显然是模最小的数的平方或者是个负数. 要求乘积最小只要求区间内最大值.最小值和绝对值小的数,再判断min,max正负就行了. 下面的代码相当于建了三个树分别存Min,Max,Mid(abs(Min)) #include<iostream> #include<cstdio> #i…

题意:平面上n个点,问你是否存在一种黑白染色方案,使得对于该方案,无法使用一条直线使得黑色点划分在直线一侧,白色点划分在另一侧.如果存在,输出一种方案. 如果n<=2,显然不存在. 如果所有点共线,且n>2,只需交替染色即可. 设凸包上的点数为K,如果K==n,且n==3,不存在,如果n>3,只需交替染色即可. 如果K<n,只需凸包上点涂黑,内部点留白即可. #include<cstdio> #include<algorithm> #include<c…

题意:给你一个序列(长度不超过2^17),支持两种操作:单点修改:询问区间中最小的ai*aj是多少(i可以等于j). 只需要线段树维护区间最小值和最大值,如果最小值大于等于0,那答案就是minv*minv: 如果最大值小于等于零,那么答案就是maxv*maxv: 要是最小值小于零,最大值大于零,答案就是minv*maxv. #include<cstdio> #include<cstring> #include<algorithm> using namespace std…

题意:给你一个串,仅含有a~g,且每个字母只出现最多一次.和一个光标初始位置,以及一个目标串,问你最少要多少的代价变化成目标串. 有五种操作:在光标前添加一个未出现过的字母,代价1. 删除光标前或者光标后的字母,代价1. 光标左移或者右移,代价0.5. 哈希,把串弄成一个八进制数,加上一个光标位置,状态数不超过8^8. 直接跑dijkstra即可. 要注意初始化的时候,可以单独记一个数组,表示用过的状态,仅仅重置这些状态,防止初始化复杂度过高. #include<cstdio> #includ…

#1586 : Minimum Time Limit:1000ms Case Time Limit:1000ms Memory Limit:256MB Description You are given a list of integers a0, a1, …, a2^k-1. You need to support two types of queries: 1. Output Minx,y∈[l,r] {ax∙ay}. 2. Let ax=y. Input The first line is…

有时候,很简单的模板题,可能有人没有做出来,(特指 I ),到时候一定要把所有的题目全部看一遍 目录 B 题解 E F 题解 H I 题解&代码 J B 输入样例 3 2 1 2 1 2 3 1 输出样例 1 说明 In the first sample case, the sequence [a1,a2,a3]=[0,1,0][a_1,a_2,a_3]=[0,1,0][a1,a2,a3]=[0,1,0] meets all the constraints and has the minimum…

这个专栏开始介绍一些<ACM国际大学生程序设计竞赛题解>上的竞赛题目,读者可以配合zju/poj/uva的在线测评系统提交代码(今天zoj貌似崩了). 其实看书名也能看出来这本书的思路,就是一本题解书,简单暴力的通过题目的堆叠来提升解决编程问题的能力. 那么下面开始探索吧. poj1037: Description Background For years, computer scientists have been trying to find efficient solutions to…

题目链接 2018 ACM 国际大学生程序设计竞赛上海大都会 下午午休起床被同学叫去打比赛233 然后已经过了2.5h了 先挑过得多的做了 .... A题 rand x*n 次点,每次judge一个点位端点的共线向量数判断是否大于给定x 强行rand 500次 代码 #include<bits/stdc++.h> using namespace std; inline int read() { int x = 0,f = 1; char c = getchar(); while(c <…

链接:https://www.nowcoder.com/acm/contest/163/F 来源:牛客网 2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it 时间限制:C/C++ 3秒,其他语言6秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 There is a matrix A that has N rows and M columns. Each grid (i,j)(0 ≤ i < N,…

2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it (扫描线) 链接:https://ac.nowcoder.com/acm/contest/163/F来源:牛客网 时间限制:C/C++ 3秒,其他语言6秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 There is a matrix A that has N rows and M columns. Each grid (i,j)(0 ≤ i…

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 (数位DP) 链接:https://ac.nowcoder.com/acm/contest/163/J?&headNav=acm来源:牛客网 时间限制:C/C++ 8秒,其他语言16秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 NIBGNAUK is an odd boy and his taste is strange a…

传送门:2018 ACM 国际大学生程序设计竞赛上海大都会赛 2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛2018-08-05 12:00:00 至 2018-08-05 17:00:00时长: 5小时 比赛情况 实录 难度差不多介于省赛和区域赛之间吧.开题A是计算几何,有点思路后就先放下去写签到题,B读错题WA一发,K直接套模板,然后就接着看A.之前写过类似题,没注意数据范围就头铁地交了发n3的代码,TE后才发现数据范围是之前那道十多倍,就听学长的先看D.推十分钟公式无果后打算直…

2016 ACM/ICPC Asia Regional Qingdao Online(部分题解) 5878---I Count Two Three http://acm.hdu.edu.cn/showproblem.php?pid=5878 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1287    Accepted Submissi…

题目链接:https://ac.nowcoder.com/acm/contest/163/J 题目大意:给定一个数N,求区间[1,N]中满足可以整除它各个数位之和的数的个数.(1 ≤ N ≤ 1012). 输入: 21018 输出: Case 1: 10Case 2: 12 解题思路:比较简单的一道数位dp题,因为N的范围最大可为10的十二次方,即数位和的范围为[1,108],1-108的最小公倍数很大不可求,所以我们直接暴力枚举数位和为1-108的情况,然后利用数位dp求出合法数的个数就可以了…

A链接:https://www.nowcoder.com/acm/contest/163/A Fruit Ninja is a juicy action game enjoyed by millions of players around the world, with squishy, splat and satisfying fruit carnage! Become the ultimate bringer of sweet, tasty destruction with every sl…

Pku 1143: Description Christine and Matt are playing an exciting game they just invented: the Number Game. The rules of this game are as follows. The players take turns choosing integers greater than 1. First, Christine chooses a number, then Matt ch…

pku 1107: Description Weird Wally's Wireless Widgets, Inc. manufactures an eclectic assortment of small, wireless, network capable devices, ranging from dog collars, to pencils, to fishing bobbers. All these devices have very small memories. Encrypti…

Poj1068: Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways: q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-seque…

这篇文章来介绍一些模拟题,即一类按照题目要求将现实的操作转换成程序语言. zoj1003: On every June 1st, the Children's Day, there will be a game named "crashing balloon" on TV.   The rule is very simple.  On the ground there are 100 labeled  balloons, with the numbers 1 to 100.  Afte…

题意:一个N*M的矩形,每个点初始都是白色的,有Q次操作,每次操作将以(x,y)为圆心,r为半径的区域涂成黑点.求最后剩余白色点数. 分析:对每行,将Q次操作在该行的涂色视作一段区间,那么该行最后的白色点数即列数-区间覆盖的总长度.这就转化成了扫描线的问题. #include<bits/stdc++.h> using namespace std; typedef long long LL; ; struct Circle{ LL x,y,r; }p[maxn]; struct Node{ LL…

题意: 给定一副n*m的格子图, 问从左上角的点开始往右下角滑,碰到墙壁就反弹, 碰到角落就停止, 问恰好经过一次的格子有多少个. 如图,恰好经过一次的格子有39个. 分析: 首先要引入两个概念, “路径长”,“格子数”. 路径长指的是整段路程的长度,如果走过同一个格子两次那么就算是2步. 格子数指的是整段路程经过的格子. 如果一个图是9*9(形如n*n)的, 那么就是从左上角一直到右下角, 走过的“路径长”恰好等于“格子数”,没有任何的格子被走过两次. 但对于一幅9 * 15这样不规则的图,“…

题目描述 We consider a positive integer perfect, if and only if it is equal to the sum of its positive divisors less than itself. For example, 6 is perfect because 6 = 1 + 2 + 3. Could you write a program to determine if a given number is perfect or not?…

题目描述 In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field, or, more generally, in a ring or even a semiring. The matrix product is designed for representing the…