CodeForces-740B Alyona and flowers】的更多相关文章

B. Alyona and flowers Problem Description: Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowe…
题目链接:Codeforces 451E Devu and Flowers 题目大意:有n个花坛.要选s支花,每一个花坛有f[i]支花.同一个花坛的花颜色同样,不同花坛的花颜色不同,问说能够有多少种组合. 解题思路:2n的状态,枚举说那些花坛的花取超过了,剩下的用C(n−1sum+n−1)隔板法计算个数.注意奇数的位置要用减的.偶数的位置用加的.容斥原理. #include <cstdio> #include <cstring> #include <cmath> #in…
Alyona and flowers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of…
题目描述: Alyona and a tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positive int…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The…
C. Alyona and mex time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Alyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special. Alyona is…
A. Alyona and copybooks time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output: standard output Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have eq…
C. Alyona and Spreadsheet time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit table…
题目: http://www.codeforces.com/contest/617/problem/C 自己感觉是挺有新意的一个题目, 乍一看挺难得(= =). 其实比较容易想到的一个笨办法就是:分别计算出每个点到喷泉的距离,然后分别按照距离远近排序(要用到两个数组),然后选定一个喷泉,从近到远依次选点,然后标记该点,从另一个喷泉中找到距离他最远且还没用被标记的点,这个距离加前一个喷泉的距离就是要求的答案,选最小的即可,n方的时间复杂度. 需要注意的几点: 1.         对于每个点最好用…
题目链接:http://codeforces.com/problemset/problem/682/C 题意:如果点v在点u的子树上且dist(u,v)>a[v]则u和其整个子树都将被删去,求被删去的点数. 思路:1为根节点,从1开始DFS遍历,记录距离dis为到祖宗节点的最大距离. #include<bits/stdc++.h> using namespace std; typedef long long ll; const int N=1e5+5; int a[N],num[N],a…