题目描述: F. Ant colonytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output Mole is hungry again. He found one ant colony, consisting of n ants, ordered in a row. Each ant i (1 ≤ i ≤ n) has a strength si.…
线段树求某一段的GCD..... F. Ant colony time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mole is hungry again. He found one ant colony, consisting of n ants, ordered in a row. Each ant i (1 ≤ i ≤ n)…
F. Ant colony time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mole is hungry again. He found one ant colony, consisting of n ants, ordered in a row. Each ant i (1 ≤ i ≤ n) has a strength si…
[BZOJ3872][Poi2014]Ant colony 试题描述 There is an entrance to the ant hill in every chamber with only one corridor leading into (or out of) it. At each entry, there are g groups of m1,m2,...,mg ants respectively. These groups will enter the ant hill one…
[BZOJ3872]Ant colony(二分,动态规划) 题面 又是权限题... Description There is an entrance to the ant hill in every chamber with only one corridor leading into (or out of) it. At each entry, there are g groups of m1,m2,...,mg ants respectively. These groups will ent…
3872: [Poi2014]Ant colony Time Limit: 30 Sec Memory Limit: 128 MB Description There is an entrance to the ant hill in every chamber with only one corridor leading into (or out of) it. At each entry, there are g groups of m1,m2,...,mg ants respecti…
codeforces 704B - Ant Man 贪心 题意:n个点,每个点有5个值,每次从一个点跳到另一个点,向左跳:abs(b.x-a.x)+a.ll+b.rr 向右跳:abs(b.x-a.x)+a.lr+b.rl,遍历完所有的点,问你最后的花费是多少 思路:每次选一个点的时候,在当前确定的每个点比较一下,选最短的距离. 为什么可以贪心?应为答案唯一,那么路径必定是唯一的,每个点所在的位置也一定是最短的. #include <bits/stdc++.h> using namespace…
Description There is an entrance to the ant hill in every chamber with only one corridor leading into (or out of) it. At each entry, there are \(g\) groups of \(m_{1},m_{2},...,m_{g}\) ants respectively. These groups will enter the ant hill one after…
题面 既然我们只知道最后数量为$k$的蚂蚁会在特殊边上被吃掉,不妨逆着推回去,然后到达每个叶节点的时候就会有一个被吃掉的蚂蚁的区间,然后二分一下就好啦 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ,maxx=1e9; int ant[N],deg[N],leaf[N]; *N],goal[*N]; int n,g,k,t,root,t1,t2,n1,n2,cnt;…
D. Ant on the Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for…
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=3872 从食蚁兽所在的边向叶节点推,会得到一个渐渐放大的取值区间,在叶子节点上二分有几群蚂蚁符合条件即可: 注意中途判断,如果已经超过范围就返回或者处理一下,据说会爆 long long 之类的: 而且食蚁兽所在的边的两个端点的初始值不一定是 k 和 k+1 !也要看度数! 注意统计答案的 num 也是 long long . 代码如下: #include<iostream> #inclu…
题目描述: Ciel the Commander time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has n cities connected by n - 1 undire…
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=3872 可以倒推出每个叶子节点可以接受的值域.然后每个叶子二分有多少个区间符合即可. 注意一开始的两个点不是直接是 l[ u ]=r[ u ]=lm !也要看度数的!且把那条边的两个端点分别算子树很方便. 而且过程中似乎会爆 long long ,所以如果 l[ i ] 都大于最大值就不往下算了:r[ i ]也要每次与最大值取min. 然后是和网上一份题解对拍出错却仍A了此题的代码.不想管是…
题目描述: The Brand New Function time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Polycarpus has a sequence, consisting of n non-negative integers: a1, a2, ..., *a**n*. Let's define function f(…
题目描述: Bus Number time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output This night wasn't easy on Vasya. His favorite team lost, and he didn't find himself victorious either - although he played p…
题目描述: Nick had received an awesome array of integers a=[a1,a2,…,an] as a gift for his 5 birthday from his mother. He was already going to explore its various properties but after unpacking he was disappointed a lot because the product a1⋅a2⋅…an of it…
