C. Misha and Forest Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev …
题目链接: 题目 A. Misha and Forest time limit per test 1 second memory limit per test 256 megabytes 问题描述 Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n ver…
传送门 Description Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev and sv…
[吐槽]:本来没打算写这题的题解的,但惨不忍睹得WA了13次,想想还是记录一下吧.自己的“分类讨论能力”本来就很差. 刚开始第一眼扫过去以为是LIS,然后忽略了复杂度,果断TLE了,说起来也好惭愧,也说明有时候太懒得动脑了,总是习惯利用惯性思维,这不是一件好事. [题意]:给你大小为n的整型数组a[n],求这数组的一个子串,其中最多可以修改子串中的一个数字,使得到的子串是最长的严格递增的子串,输出该子串的长度 L. [思路]:O(n)复杂度,枚举断点情况.第0个和第n个位置默认为断点.(用ve[…
A题: 贪心水题,注意1,1这组数据,坑了不少人 #include <iostream> #include <cstring> using namespace std; int main() { int a1,a2; while(cin>>a1>>a2) { int i=0; int b = max(a1,a2); int s = min(a1,a2); if(b==1 && s==1) { cout<<0<<endl…
题目链接: 传送门 Domino Effect time limit per test:1 second memory limit per test:256 megabytes Description Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix. The dot product…
题目链接: 传送门 Domino Effect time limit per test:1 second memory limit per test:256 megabytes Description Little Chris knows there's no fun in playing dominoes, he thinks it's too random and doesn't require skill. Instead, he decided to play with the…
Equal Sums time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given kk sequences of integers. The length of the ii-th sequence equals to nini. You have to choose exactly two sequences…
C题意: 给定n个点(标号0~n-1)的度数(就是与其邻接的点的个数)和所有与它邻接的点标号的异或和,求满足这些条件的树的边应该是怎么连的,将边输出出来 这里可以理解成拓扑排序的方式考虑,当i度数为1的时候,那么我们必然知道 i 肯定得与s[i]相连使其剩余的异或值为0 所以建立一个队列不断将度数变为1的点放进来,得到边以后,不断更新点的度数和其对应的异或和的值 #include <bits/stdc++.h> using namespace std; <<); int n,deg…
