Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 全场题解 菜鸡只会A+B+C,呈上题解: A. Bear and Big Brother 题意:我也没看太清,就是给你两个10以内的数a,b.a每天乘以3,b每天乘以2,求多少天后a大于b. 思路:应该是有公式的,不过看到数据这么小直接暴力乘求解.官方题解貌似就是这样,数据小就是水题. const int N=1e3+10; int main() { int a…

C. Bear and Different Names 题目连接: http://codeforces.com/contest/791/problem/C Description In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should s…

B. Bear and Friendship Condition 题目连接: http://codeforces.com/contest/791/problem/B Description Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pic…

我们要统计的答案是sigma([L/K]),L为路径的长度,中括号表示上取整. [L/K]化简一下就是(L+f(L,K))/K,f(L,K)表示长度为L的路径要想达到K的整数倍,还要加上多少. 于是,我们现在只需要统计sigma((L+f(L,K))),最后除以K即可. 统计sigma(L)时,我们考虑计算每条边出现在了几条路径中,设u为edgei的子节点,那么这条边对答案的贡献就是siz(u)*(n-siz(u)),siz(u)为u的子树大小. 统计sigma(f(L,K))时,我们需要dp出…

A 模拟 B 发现对于每个连通块,只有为完全图才成立,然后就dfs C 构造 想了20分钟才会,一开始想偏了,以为要利用相邻NO YES的关系再枚举,其实不难.. 考虑对于顺序枚举每一个NO/YES,与前一个需要用的的字符串有k-1个交集,只多了一个string 于是只要保证k-1个string不同,只通过当前的string来影响答案, 若YES,开一个新的串: NO,则和第一个串一样,以此让这个相同串在下一次枚举中消失.. //怎么想到? 其实NO可能有多个相同名字,关系就很乱了,我们希望只有…

A. Bear and Big Brother time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak…

如果某个位置i是Y,直接直到i+m-1为止填上新的数字. 如果是N,直接把a[i+m-1]填和a[i]相同即可,这样不影响其他段的答案. 当然如果前面没有过Y的话,都填上0就行了. #include<cstdio> #include<string> #include<iostream> using namespace std; int n,m,a[60],e; string ma[60]; int main(){ // freopen("a.in",…

Description Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to re…

Description A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them. Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 throu…

Description In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if…

Description Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures). There are n members, numbered 1 through n. m pairs of members are friends.…

Description Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh a and b respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.…

A. Vicious Keyboard time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string s with only t…

地址:http://codeforces.com/contest/801/problem/D 题目: D. Volatile Kite time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a convex polygon P with n distinct vertices p1, p2, ..., p…

地址:http://codeforces.com/contest/801/problem/C 题目: C. Voltage Keepsake time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have n devices that you want to use simultaneously. The i-th devi…

A. Vicious Keyboard 题意:给你一个字符串,里面只会包含VK,这两种字符,然后你可以改变一个字符,你要求VK这个字串出现的次数最多. 题解:数据范围很小,暴力枚举改变哪个字符,然后check就好. #include<bits/stdc++.h> using namespace std; int main(){ string s; cin>>s; int ans = 0; for(int i=0;i<s.size();i++){ if(s[i]=='V'){…

A 每次可以换一个或不换,暴力枚举位置即可 B 模拟 C 二分答案.. 边界可以优化r=totb/(tota-p),二分可以直接(r-l>=EPS,EPS不要太小,合适就好),也可以直接限定二分的次数(防止死循环! 一开始EPS=1e-10,结果TLE) 注意精度问题 long double,在统计和的时候注意long long…

A. Vicious Keyboard time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Tonio has a keyboard with only two letters, "V" and "K". One day, he has typed out a string s with only t…

A. Is it rated? time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to a…

A. Is it rated? time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to a…

http://codeforces.com/contest/967/problem/F 题目大意: 有n个点,n*(n-1)/2条边的无向图,其中有m条路目前开启(即能走),剩下的都是关闭状态 定义:从x走到y(即x->y)后,和x所连接的边的所有状态都反转(即开启->关闭,关闭->开启) 问,从起点1走到终点n,最少需要经过几步,并且输出这个路径(如果存在多挑最短路径,输出任意一条) 如果不存在,则输出-1 思路: 其实这道题第一眼看过去就是bfs,不过,由于每个状态都在改变,那要怎么…

http://codeforces.com/contest/967/problem/E 题目大意: 给你一个数组a,a的长度为n 定义:b(i) = a(1)^a(2)^......^a(i), 问,是否存在一种情况,使得a(i)重新排列以后以后,满足b(i)是严格单调递增的. 思路: 对于二进制位操作来说,如果异或后值增加,即 p^q > p,必然满足两种情况(假定一个为p,一个为q) ①q的最高位>p的最高位 ②q的最高位(假定第k为最高) < p的最高位,但是p的第k位为0 这样,…

http://codeforces.com/contest/967/problem/D 题目大意: 有n个服务器,标号为1~n,每个服务器有C[i]个资源.现在,有两个任务需要同时进行,令他为x1,x2. 运行任务的条件: ①每个服务器只能同时运行一个任务 ②任务可以同时分配到多个服务器中执行.假设任务x1分配到a个服务器中,则每个服务器都需要使用x1/a的资源(该资源可以为小数) 问能否满足以上条件,使得x1,x2同时在服务器中运行? 思路: 将所有服务器按照资源增序排列,定义dp(i)表示从…

http://codeforces.com/contest/948/problem/A   A. Protect Sheep Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his she…

题: OvO http://codeforces.com/contest/947/problem/D 923D 947D 948E 解: 记要改变的串为 P1 ,记目标串为 P2  由变化规则可得: 1. B -> AC -> AAB -> AAAC -> C ( 即 B -> C -> B ) 2. AB -> AAC -> AAAB -> B (即 AB -> B ) 3. B -> AC -> AB ( 即 B -> AB…

题目链接  题意  每天有体积为Vi的一堆雪,所有存在的雪每天都会融化Ti体积,求出每天具体融化的雪的体积数. 分析 对于第i天的雪堆,不妨假设其从一开始就存在,那么它的初始体积就为V[i]+T[1..i-1],在第i天则需要融化T[i]体积,若T[1....i]>=V[i]+T[1...i-1],那么这堆雪就融化完了,融化的体积为V[i]+T[1..i-1] - T[1...i-1]:否则就为T[i].用个优先队列来维护,由于默认是数值大的优先,所以实际处理中添加一个负号. #include<…

题意:有两个服务要求被满足,服务S1要求x1数量的资源,S2要求x2数量的资源.有n个服务器来提供资源,第i台能提供a[i]的资源.当你选择一定数量的服务器来为某个服务提供资源后,资源需求会等量地分担给它们,要求每台服务器承担的资源需求不超过其所能提供的资源需求.给定一种合法的方案,每台服务器要么没有被分配给任何一个服务,或者被分配给其中一个服务. 对服务器按能提供的资源从小到大排序.枚举给S1分配的服务器数量i,然后在a数组中二分,就可以得到给S1提供的是哪i台服务器,它们占据了a数组中连续的…

题意:海平面每天高度会变化,一个人会在每天海平面的位置刻下一道痕迹(如果当前位置没有已经刻划过的痕迹),并且记录下当天比海平面高的痕迹有多少条,记为a[i].让你最小化每天比海平面低的痕迹条数之和. 贪心,使得痕迹条数尽可能小. 考虑处理出每天痕迹条数的数组f[i]的最小可能值. 满足几个约束即可,f[i]>=f[i-1],f[i]>=a[i]+1,f[i]>=f[i+1]-1.前后扫两遍就可以处理出f数组,然后就可以轻松统计答案了. #include<cstdio> #in…

题意:给你一个棋盘的最终局面. 你的一次操作可以选择一些行和列,将它们的交叉点染黑,不能重复选择某行或者某列.问你是否能经过数次操作之后,达到目标局面. 就枚举所有黑点,如果该点行列都没被标记,就给它的行列新建一次操作的序号:否则如果只有行有操作标号,就把它付给列操作标号:对列亦然. 然后模拟一遍,看看结果是否和题目所给相同. #include<cstdio> #include<cstdlib> using namespace std; int n,m,num; char a[55…

A. Protect Sheep time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus dec…