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前置扯淡 %%@\(wucstido\),思路是在是巧妙---link Description 给一个长度为\(n\)由 \(<\) 和 \(>\)组成的字符串,表示序列中相邻位置的数的大小关系 求构造两种排列,使得其在满足上述条件的情况下的 \(LIS\) 分别最长\(/\)最短 Solution 直接考虑构造最短的排列,然后最长的同理(真的是把符号改改就可以了) 整个排列\(LIS\)的最小值在最长的一段\("<"\)中取得,然后我们考虑构造 先把整个排列逆序过来…
Codeforces Round #620 (Div. 2) D. Shortest and Longest LIS 题解: 贪心即可,对于最短序列,我们尽可能用可用的最大数字放入序列中,对于最长序列,我们尽可能用可用的最小数组放入序列即可,再处理序列时,当满足当前防止变化规律的符号直接防止,如果不满足则向后遍历直到遇到满足条件的符号,然后倒序放直到放完.eg. "><<>" 我们从大向小取,第一个为大于号所以直接填入,第二为小于号暂时不填,第三为小于号暂时不填…
看样例,>><>><,要构造 LIS 最短的,我们需要找最小链划分的方案,即包含最少的下降列 很容易想到把连续 < 的看成一段,比如样例就是 .|.|. .|.|. . 每一段内必须上升,考虑让它连续,然后让段末取当前没取过的最大值即可 要构造 LIS 最长的,同理,我们把连续 > 的堪称一段,比如样例就是 . . .|. . .|. 每段内必须下降,考虑让它连续,然后让段膜取当前没去过的最小值即可 #include <bits/stdc++.h>…
根据题目,我们可以找最短的LIS和最长的LIS,找最短LIS时,可以将每一个increase序列分成一组,从左到右将最大的还未选择的数字填写进去,不同组之间一定不会存在s[i]<s[j]的情况,保证满足题意,找最长LIS,可以找补集,将每个decrease序列分成一组,找到后取反即可 #include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; vec…
https://codeforces.com/contest/1304/problem/D #include<bits/stdc++.h> using namespace std; void solve(){ int n; string s; cin>>n; cin>>s; int Min[n]; int t = n; ;i<s.length();i++){ ,indx = i; if(s[i] == '<'){ while(s[i]=='<'){ l…
A. Two Rabbits (手速题) #include<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ int t; cin>>t; while(t--){ ll x,y,a,b; cin>>x>>y>>a>>b; ll sum = a+b; ){ cout<<(y-x)/sum<<endl; } else{ cou…
A. Two Rabbits 思路: 很明显,如果(y-x)%(a+b)==0的话ans=(y-x)/(a+b),否则就为-1 #include<iostream> #include<algorithm> using namespace std; +; int main(){ int k; scanf("%d",&k); while(k--){ long long x,y,a,b,ans; cin>>x>>y>>a&g…
Codeforces Round #620 (Div. 2) A. Two Rabbits 题意 两只兔子相向而跳,一只一次跳距离a,另一只一次跳距离b,每次同时跳,问是否可能到同一位置 题解 每次跳相对距离减少Δ=(a+b)\Delta=(a+b)Δ=(a+b),如果总距离是Δ\DeltaΔ的倍数,就证明可以到同一位置 代码 #include<iostream> #include<cstdio> using namespace std; int main() { long N,x…
原题链接在这里:https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/ 题目: Given an unsorted array of integers, find the number of longest increasing subsequence. Example 1: Input: [1,3,5,4,7] Output: 2 Explanation: The two longe…
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Lweb and String Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 368    Accepted Submission(s): 243 Problem Description Lweb has a string S. Oneday, he decided to transform this string to a new s…
Lweb and String 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5842 Description Lweb has a string S. Oneday, he decided to transform this string to a new sequence. You need help him determine this transformation to get a sequence which has the longe…
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Problem Description Lweb has a string $S$. Oneday, he decided to transform this string to a new sequence. You need help him determine this transformation to get a sequence which has the longest LIS(Strictly Increasing). You need transform every lette…
Problem Description Lweb has a string S. Oneday, he decided to transform this string to a new sequence. You need help him determine this transformation to get a sequence which has the longest LIS(Strictly Increasing). You need transform every letter…
Lweb and String Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 492    Accepted Submission(s): 312 Problem Description Lweb has a string S. Oneday, he decided to transform this string to a new s…
题目链接 题目要求: Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. For example: Given "aacecaaa", return "a…
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Lweb and String 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5842 Description Lweb has a string S. Oneday, he decided to transform this string to a new sequence. You need help him determine this transformation to get a sequence which has the longe…
Lweb and String Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 261    Accepted Submission(s): 174 Problem Description Lweb has a string S. Oneday, he decided to transform this string to a new s…
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Longest Increasing Subsequence(LIS) 一个美丽的名字 非常经典的线性结构dp [朴素]:O(n^2) d(i)=max{0,d(j) :j<i&&a[j]<a[i]}+1 直接两个for [二分查找优化]:O(n^2) g(i):d值为i的最小的a  每次更新然后lower_bound即可 [大于等于] lower_boundReturn iterator to lower bound Returns an iterator pointing…
题目传送门 题意:LIS(Longest Increasing Subsequence)裸题 分析:状态转移方程:dp[i] = max (dp[j]) + 1   (a[j] < a[i],1 <= j < i) 附带有print输出路径函数 代码: #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N = 1e4 + 10…
传送门 The task is to find the length of the longest subsequence in a given array of integers such that all elements of the subsequence are sorted in ascending order. For example, the length of the LIS for { 15, 27, 14, 38, 26, 55, 46, 65, 85 } is 6 and…
Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 47465   Accepted: 21120 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ...…