题目:click here #include <bits/stdc++.h> using namespace std; typedef unsigned long long ll; const int INF = 0x3f3f3f3f; ; int n; int a[M][M]; // a[i][j] 表示从[i][0]到[i][j]的和 int main() { while( ~scanf("%d", &n ) ) { memset( a, , sizeof(a)…

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[…

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…

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…

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…

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 problem the sub-rectangle with the largest sum is referred to…

点我看题目 题意 : 给你一个n*n的矩阵,让你找一个子矩阵要求和最大. 思路 : 这个题都看了好多天了,一直不会做,今天娅楠美女给讲了,要转化成一维的,也就是说每一列存的是前几列的和,也就是说 0 -2 -7 0 9 2 -6 2-4 1 -4 1-1 8 0 -2 处理后就是:0  -2  -9  -99   11  5   7-4 -3  -7  -6-1  7   7   5 #include <iostream> #include <stdio.h> #include &…

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…

题目链接:https://codeforces.com/contest/1373/problem/D 题意 给出一个大小为 $n$ 的数组 $a$,下标为 $0 \sim n - 1$,可以进行一次反转一个区间中元素的操作,问偶数下标元素的最大和, 题解 如果反转区间长度为奇数,则下标奇偶性不同的元素间不会互换,所以反转的区间长度为偶数,反转后的区间可以看作相邻元素两两交换所得. 如:1 2 3 4 反转后为 4 3 2 1,偶数下标元素由 1 3 变成了 2 4 ,可以看作 1 与 2 相交换…

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…