“Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.” 我的代码是这样的: /* * @lc app=leetcode.cn id=491 lang=cpp * * [491] Non-decreasing Subsequences */#...
题意: 戳这里查看 分析: 我们先不考虑区间的限制设出DP状态,$f[i][j]$表示枚举到第$i$个数,单调不降序列最后一位是$j$的方案数 转移方程就是: if(a[i]!=j) f[i][j]=f[i-1][j] else f[i][j]=f[i-1][k](k<=j) 我们发现可以用矩
Given an integer arraynums, returnall the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer inany order. Example 1: Input:nums = [4,6,7,7]Output:[[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7...
Reverse mathematicsEvery function over the natural numbers has an infinite subdomain on which the function is non-decreasing. Motivated by a question of Dzhafarov and Schweber, we study the reverse mathematics of var...doi:10.1007/s00153-017-0536-9Ludovic Patey...
[loj3247]Non-Decreasing Subsequences,分治,考虑分治到[l,r]的区间(设$mid=(l+r)/2$),将询问分为两类:1.在左/右区间,直接递归下去;2.跨越中间,那么处理出两个数组:L[i][j]表示左区间在开头第i个位置,以数字j为结尾的上升子序列个数(不跨越mid),右区间同理(L
Luogu P6009 [USACO20JAN]Non-Decreasing Subsequences P Link 先考虑计算\(l=1,r=n\)时的答案。 很显然我们可以dp,设\(f_{i,j}\)表示考虑前\(i\)个数,NDS末尾为\(j\)的方案数,那么转移为: \[f_{i,j}= \begin{cases} f_{i-1,j}&j\ne a_i\\ f_{i-1,j}+\sum\limits_{l=1}^{...
Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order. Example 1: Input: nums = [4,3,10,9,8]Output: [10,9] Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements...
Now it is possible to pass directly to code of the main function of sorting by natural merging with the non-decreasing and decreasing arranged subsequences. Code of the Main Function NaturalMergeSort C++ Shrink ▲ /// // NaturalMergeSort /// BOOL NaturalMergeSort(NMS *pNms) { //*** ...
increasing-decreasing-string.c increasing-order-search-tree.c increasing-subsequences.c increasing-triplet-subsequence.c insert-delete-getrandom-o1-duplicates-allowed.c insert-delete-getrandom-o1.c insert-interval.c insert-into-a-binary-search-tree.c insertion-sort-list.c insufficient-no...
题面https://loj.ac/problem/3247 题解 考虑CDQ分治。对于$solve(l,r,v)$(其中l,r表示当前处理到的区间的左右端点,v是一个vector,存放当前区间内待处理的所有询问的序号):设$m=(l+r) 1$。 预处理出$f "i][j"