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;…

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…

[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…

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]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…

[BZOJ3872][Poi2014]Ant colony Description 给定一棵有n个节点的树.在每个叶子节点,有g群蚂蚁要从外面进来,其中第i群有m[i]只蚂蚁.这些蚂蚁会相继进入树中,而且要保证每一时刻每个节点最多只有一群蚂蚁.这些蚂蚁会按以下方式前进: ·在即将离开某个度数为d+1的点时,该群蚂蚁有d个方向还没有走过,这群蚂蚁就会分裂成d群,每群数量都相等.如果d=0,那么蚂蚁会离开这棵树. ·如果蚂蚁不能等分,那么蚂蚁之间会互相吞噬,直到可以等分为止,即一群蚂蚁有m只,要分成…

Ant colony 题解: 因为一个数是合法数,那么询问区间内的其他数都要是这个数的倍数,也就是这个区间内的gcd刚好是这个数. 对于这个区间的gcd来说,不能通过前后缀来算. 所以通过ST表来询问这个区间的gcd. 那么题目就变成了询问一个区间内有多少个k. 我们对于每个数都离散化之后,在相应的数字存下下标. 然后二分一下个数. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.…

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.…