~~~题面~~~ 题解: 观察到拼接后的数据范围过大,无法O(n)解决,但是大区间是由很多小区间组成,而小区间是固定的,不会变化,所以可以考虑预处理出每个小区间的信息,然后根据给定序列按顺序一步一步合并区间信息. 跟线段树维护区间最大子段和类似,要合并2个区间我们只需要知道如下信息: 前缀最大子段和,后缀最大子段和,当前区间最大子段和. 那么转移方式如下: 1,对于区间最大子段和而言,要么是继承2个区间中的某个最大子段和,要么是用上一个区间的后缀最大子段和+后一个区间的前缀最大子段和凑成一个子段…

B. Maximum Sum of Digits time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output You are given a positive integer nn. Let S(x)S(x) be sum of digits in base 10 representation of xx, for example, S(…

C. Little Girl and Maximum Sum time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The little girl loves the problems on array queries very much. One day she came across a rather well-known proble…

Maximum sum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 39599   Accepted: 12370 Description Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: Your task is to calculate d(A). Input The input consists o…

1146. Maximum Sum Time limit: 0.5 secondMemory limit: 64 MB Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this p…

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=44  Maximum Sum  Background A problem that is simple to solve in one dimension is often much more difficult to solve in more th…

题目传送门 /* 最大子矩阵和:把二维降到一维,即把列压缩:然后看是否满足最大连续子序列: 好像之前做过,没印象了,看来做过的题目要经常看看:) */ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int dp[…

Maximum Sum 大意:给你一个n*n的矩阵,求最大的子矩阵的和是多少. 思路:最開始我想的是预处理矩阵,遍历子矩阵的端点,发现复杂度是O(n^4).就不知道该怎么办了.问了一下,是压缩矩阵,转换成最大字段和的问题. 压缩行或者列都是能够的. int n, m, x, y, T, t; int Map[1010][1010]; int main() { while(~scanf("%d", &n)) { memset(Map, 0, sizeof(Map)); for(i…

1146. Maximum Sum Time limit: 1.0 second Memory limit: 64 MB Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this…

题目大意:UVa 108 - Maximum Sum的加强版,求最大子矩阵和,不过矩阵是可以循环的,矩阵到结尾时可以循环到开头.开始听纠结的,想着难道要分情况讨论吗?!就去网上搜,看到可以通过补全进行处理,也是,通过补全一个相同的,问题就迎刃而解了,所以把n*n的矩阵扩展成2n*2n的矩阵就好了. #include <cstdio> #include <cstring> #define MAXN 160 int a[MAXN][MAXN], sum[MAXN][MAXN]; int…