A,有多个线段,求一条最短的线段长度,能过覆盖到所又线段,例如(2,4)和(5,6) 那么我们需要4 5连起来,长度为1,例如(2,10)(3,11),用(3,10) 思路:我们想一下如果题目说的是最长我们肯定是取最小x和最大的y连起来就完事. 但是要求长度最小又得覆盖,那么可以这样想,我们需要把最小的x不断右移到这条线段的y, 最大的左移到x,所以就是最大x-最小y完事 #include <bits/stdc++.h> using namespace std; #define ll long…

F2. Wrong Answer on test 233 (Hard Version) Your program fails again. This time it gets "Wrong answer on test 233" . This is the harder version of the problem. In this version, 1≤n≤2⋅105. You can hack this problem if you locked it. But you can h…

E. Arson In Berland Forest The Berland Forest can be represented as an infinite cell plane. Every cell contains a tree. That is, contained before the recent events. A destructive fire raged through the Forest, and several trees were damaged by it. Pr…

D2. Optimal Subsequences (Hard Version) This is the harder version of the problem. In this version, 1≤n,m≤2⋅105. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems. You are given a seque…

C. Messy You are fed up with your messy room, so you decided to clean it up. Your room is a bracket sequence s=s1s2-sn of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'. In one operation you can choo…

B. Box Permutation p is a sequence of integers p=[p1,p2,-,pn], consisting of n distinct (unique) positive integers between 1 and n, inclusive. For example, the following sequences are permutations: [3,4,1,2], [1], [1,2]. The following sequences are n…

A. Math Problem Your math teacher gave you the following problem: There are n segments on the x-axis, [l1;r1],[l2;r2],-,[ln;rn]. The segment [l;r] includes the bounds, i.e. it is a set of such x that l≤x≤r. The length of the segment [l;r] is equal to…

//因为可以反转n次 所以可以得到任何可以构成的序列 #include<iostream> #include<string> #include<vector> using namespace std ; typedef pair<int,int>PII; int n,k; string s; string get_str(int n,int k) {//先构建前k-1个 string res=""; ; i<k-; i++) { r…

#include<bits/stdc++.h> using namespace std; ]; ]; int main() { int total; cin>>total; while(total--) { int n; bool flag=true; cin>>n; unordered_set<int>tree; set<int>num; ; i<=n; i++) num.insert(i);//插入每个数字 从小到大 ; i<=n…

//只要从所有区间右端点的最小值覆盖到所有区间左端点的最大值即可 #include<iostream> using namespace std ; int x,y; int n; int t; int main() { cin>>t; while(t--) { cin>>n; ) { cin>>x>>y; cout<<<<endl; } else { ,r=1e9+; ; i<=n; i++) { cin>&g…