题意:给你一棵树,每个点有一个左括号或者右括号,问你树上能够完美匹配的路径数量(l->r,r->l 视作不同路径). 点分治可以使用“不扣去重复答案”的写法,只不过,要先将每个点的子树按照从小到大的顺序排序,防止复杂度出错.(此题不需要,因为统计一个子树的贡献的时候,时间复杂度最多只与当前子树的大小有关,与之前的无关) 要将重心归到“之前的所有子树中”,而“当前的子树”不含重心. 把左括号视作1,右括号视作-1. 对于每颗子树中的点,统计三个数组,分别表示重心下面的结点到当前结点的路径的前缀和…

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=648&page=show_problem&problem=5159 In normal football games, the winner team gets 3 points, loser team gets 0 point, and if there is a drawgame, bot…

Chef and Apple Trees Chef loves to prepare delicious dishes. This time, Chef has decided to prepare a special dish for you, and needs to gather several apples to do so. Chef has N apple trees in his home garden. Each tree has a certain (non-zero) num…

题目:Problem A. Arithmetic DerivativeInput file: standard inputOutput file: standard inputTime limit: 1 secondMemory limit: 256 mebibytesLets define an arithmetic derivative:• if p = 1 then p0 = 0;• if p is prime then p0 = 1;• if p is not prime then n0…

题目:Problem D. Clones and TreasuresInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 256 mebibytesThe magical treasury consists of n sequential rooms. Due to construction of treasury its impossible togo from room…

题目: Problem D. Great AgainInput file: standard inputOutput file: standard outputTime limit: 2 secondsMemory limit: 512 megabytesThe election in Berland is coming. The party United Berland is going to use its influence to win themagain. The crucial co…

题目:Problem K. PiecemakingInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 512 mebibytesThe civil war in Berland continues for five years already. The United Nation decided to end the bloodshed.Berland consists o…

题意:给你一张简单无向图(但可能不连通),再给你一个K,让你求解任意一个问题:K染色或者输出一条K长路径. 直接贪心染色,对一个点染上其相邻的点的颜色集合之中,未出现过的最小的颜色. 如果染成就染成了.如果到某个点,发现染不成,则倒着按照颜色从大到小回去,则一定恰好可以找出一条K长度的路径. #include<cstdio> #include<cstring> using namespace std; int first[1005],next[20005],v[20005],e,c…

题意:有n个土豆,每个有体积V(i),你可以将每个土豆等分为不超过K份,问你最大块和最小块比值最小为多少. 直接枚举切法,只有n*K种,然后保证其为最大块,去算其他块的切法,即让其他块切得尽可能大即可.O(n*n*K). #include<cstdio> #include<cstring> using namespace std; int n,K,a[105],ans1=150,ans2=1,x[105],ansx[105]; int main(){ // freopen(&quo…

题意: 给你一个n,问你R(n)对应的字符串长度最小的是啥. dp打个表出来,f(i)表示i值对应的字符串的最小长度,发现f(1)=1,f(2)=2,其他的情况下,若是偶数,则恰好在其外面加一对中括号,然后中间填i/2最优,若是奇数,恰好在i-1前面加个1最优. 于是递归输出答案即可. #include<cstdio> #include<iostream> #include<string> using namespace std; string work(int x){…