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D. Progressions Covering_线段树维护差分
线段树维护差分 D. Progressions Covering 题目大意： 数列a原来全是0，可以无限次进行一种操作，每次操作可以选择一段长度为k的区间，对该区间的数字分别对应加上1,2,3,...,k。再给出数列b，问最少操作几次可以使得a数列的每一个数字不小于b数列中的对应数字。 思路和代码： 可以操作题目给Working with covering indexes
Working with covering indexes As you could read in the indexing basics article, a well-defined index can boost query performance, but there are a few more basic tricks that can have a great impact on how your query is executed. One of the most important i做题集——(矩阵快速幂+推公式)Covering
题目： http://acm.hdu.edu.cn/showproblem.php?pid=6185 题面： Sample Input 1 2 Output 1 5 题意： 有地毯为2 * 1。有操场4 * n，地毯尽可以横放及竖放，求地毯不重叠的铺满整个地板有多少种方法。 思路： 多少种方法其实可以联想到dp，每一个n也刚好可以对应一种状态，但是吴永辉教授2021年讲课3-4
目录 B Coin Toss B Coin Toss 戳我. Description In a popular carnival game, a coin is tossed onto a table with an area that is covered with square tiles in a grid. The prizes are determined by the number of tiles covered by the coin when it comes toLeetCode 632. Smallest Range Covering Elements from K Lists
原题链接在这里：https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/ 题目： You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the ranCodeChef - COVERING Covering Sets（sosdp）
链接 题面： 定义：R(x)=∑x&(i∣j∣k)=xA[i]∗B[j]∗C[k]R(x)=\sum_{x\&(i|j|k)=x}A[i]*B[j]*C[k]R(x)=∑x&(i∣j∣k)=xA[i]∗B[j]∗C[k] 题目就变成了∑x∈[0,2n)R(x)\sum_{x\in[0,2^{n})}R(x)∑x∈[0,2n)R(x) 思路： 先用sosdp从前往后转移，求出A、B、C的状态，然