Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 251171    Accepted Submission(s): 59503 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 206582    Accepted Submission(s): 48294 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 81735    Accepted Submission(s): 18797 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

这个题对我来说真的是相当难的题目了,严格来讲可能不算是个动态规划的题目,但这个题目对类似的划分多个非重叠连续子区间的问题提供了一个很好解决方案 这个题目需要找三个非重叠的连续子区间,通过维护两个数组将第一个和第三个子区间可能的开始pos记录下来,在中间那个子区间开始的pos遍历时限制其边界范围,根据长度能恰到好处的将三个子区间划分成非重叠的,还使用了集合相减代替累加这样比较简单好用的技巧 下面提供代码: class Solution { public int[] maxSumOfThreeSub…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 206582    Accepted Submission(s): 48294 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

2018-07-12 23:21:53 问题描述: 问题求解: dp[i][j] : 以ai结尾的分j个部分得到的最大值 dp[i][j] = max{dp[k][j - 1] + (ak+1 + ... + ai) / (i - k)} k = [j - 2, i - 1] public double largestSumOfAverages(int[] A, int K) { double[][] dp = new double[A.length][K + 1]; int curSum =…

时间复杂度O(n) 空间复杂度O(1) #include <cstdio> int main() { int T;scanf("%d",&T); ,n,a,l,r,s,L,R,S;t<=T;t++) { scanf(; ;r<=n;r++) { scanf(?s=S=a:s+=a; if(s>S)S=s,L=l,R=r; )s=,l=r+; } printf("Case %d:\n%d %d %d\n",t,S,L,R); if…

Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. I…

[题目] 把N个骰子扔在地上,所有骰子朝上一面的点数之和为S.输入N,打印出S的所有可能的值出现的概率. [分析] 典型的动态规划题目. 设n个骰子的和为s出现的次数记为f(n,s),其中n=[1-N],s=[n-6n]. n=1, s=[1-6], f(n,s)=1; n=[2-N], s=[n-6n], f(n,s)= f(n-1,s-1)+ f(n-1,s-2)+ f(n-1,s-3)+ f(n-1,s-4)+ f(n-1,s-5)+ f(n-1,s-6) = sum(f(n-1,s-t)…

http://acm.hdu.edu.cn/showproblem.php?pid=1003 给出一个包含n个数字的序列{a1,a2,..,ai,..,an},-1000<=ai<=1000 求最大连续子段和及其起始位置和终止位置,很基础的动态规划(DP)问题,看完DP第一次做的DP题目 DP真的是一种很优美的算法,或者说思想,但是比较难理解,我对DP的理解还很浅薄 # include <stdio.h> # define INF 1000000000 int main() { i…

这是一道水题,作为没有货的水货楼主如是说. 题意:已知一个数组nums {a1,a2,a3,.....,an}(其中0<ai <=1000(1<=k<=n, n<=20))和一个数S c1a1c2a2c3a3......cnan = S, 其中ci(1<=i<=n)可以在加号和减号之中任选. 求有多少种{c1,c2,c3,...,cn}的排列能使上述等式成立. 例如: 输入:nums is [1, 1, 1, 1, 1], S is 3. 输出 : 5符合要求5种…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 250714    Accepted Submission(s): 59365 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

题意大概就是,给定一个包含非负整数的序列nums以及一个整数m,要求把序列nums分成m份,并且要让这m个子序列各自的和的最大值最小(minimize the largest sum among these m subarrays). Note:If n is the length of array, assume the following constraints are satisfied: 1 ≤ n ≤ 10001 ≤ m ≤ min(50, n) Examples: Input:num…

HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For exampl…

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 193355 Accepted Submission(s): 45045 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…

Unique Paths https://oj.leetcode.com/problems/unique-paths/ A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to rea…

很久之前看过这个题目,但是没有仔细整理,直到现在看基础才想到这两个题.这两个题非常经典也非常类似.接下来分别介绍. 部分和问题 题目描述 给定整数a1.a2........an,判断是否可以从中选出若干数,使它们的和恰好为K. 输入 首先,n和k,n表示数的个数,k表示数的和.接着一行n个数.(1<=n<=20,保证不超int范围) 输出 如果和恰好可以为k,输出“YES”,并按输入顺序依次输出是由哪几个数的和组成,否则“NO” 样例输入 4 13 1 2 4 7 样例输出 YES 2 4 7…

引言 二维动态规划中最常见的是棋盘型二维动态规划. 即 func(i, j) 往往只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有关 这种情况下,时间复杂度 O(n*n),空间复杂度往往可以优化为O(n) 例题  1 Minimum Path Sum  Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whi…

问题描述:Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. 问题分析:参考路径问题,其实就是加权的路径问题,求最小的权值和.动态规划问题,核心递推公式,d[i][j] = min(d[i-1][j],d[i][j-1])+a[i][j]. publi…

原创 http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目要求求出一个序列里面的最大序列和,序列要求是连续的,给出最大序列和,序列首元素下标和尾元素下标,按特定的格式输出. 解题思路: 动态规划,我们可以将所有序列按以序列中的元素a[i](i=1~n)结尾进行分类,比如: 以a[1]结尾的序列有:a[1] 以a[2]结尾的序列有:a[1]a[2],a[2] 以a[3]结尾的序列有:a[1]a[2]a[3],a[2][3],a[3] ......  这…

Leetcode 931. Minimum falling path sum 最小下降路径和(动态规划) 题目描述 已知一个正方形二维数组A,我们想找到一条最小下降路径的和 所谓下降路径是指,从一行到下一行,只能选择间距不超过1的列(也就是说第一行的第一列,只能选择第二行的第一列和第二列:第二行的第二列,只能选择第三行的第一列第二列第三列),最小下降路径就是这个路径的和最小 测试样例 Input: [[1,2,3],[4,5,6],[7,8,9]] Output: 12 Explanation:…

python编写PAT甲级 1007 Maximum Subsequence Sum wenzongxiao1996 2019.4.3 题目 Given a sequence of K integers { N​1, N2, ..., N​K}. A continuous subsequence is defined to be { N​i, N​i+1, ..., N​j} where 1≤i≤j≤K. The Maximum Subsequence is the continuous sub…

Max Sum Plus Plus 题意:题意理解了老半天,这里是说在给定数列中,取m组子数列,不能有重复,使得这些子序列的和最大: 就比如m=2时候,1 /2/-4/5/6.可以不用拿-4的意思: 思路:这道题的思路是动态规划,递推: 状态dp[i][j] 表示有前j个数,组成i组的和的最大值. 决策: 第j个数,要么包含在第i组里面,要么自己独立成组. 其中最后一组包含a[j].(这很关键) 则状态转移方程为:(在二维图中,就是要么从左边取,要么取上一行的最大值,下式中,左边max是包含在第…

Leetcode之动态规划(DP)专题-931. 下降路径最小和(Minimum Falling Path Sum) 给定一个方形整数数组 A,我们想要得到通过 A 的下降路径的最小和. 下降路径可以从第一行中的任何元素开始,并从每一行中选择一个元素.在下一行选择的元素和当前行所选元素最多相隔一列. 示例: 输入:[[1,2,3],[4,5,6],[7,8,9]] 输出:12 解释: 可能的下降路径有: [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9] [2…

Leetcode之动态规划(DP)专题-712. 两个字符串的最小ASCII删除和(Minimum ASCII Delete Sum for Two Strings) 给定两个字符串s1, s2,找到使两个字符串相等所需删除字符的ASCII值的最小和. 示例 1: 输入: s1 = "sea", s2 = "eat" 输出: 231 解释: 在 "sea" 中删除 "s" 并将 "s" 的值(115)加入总…

Leetcode之动态规划(DP)专题-64. 最小路径和(Minimum Path Sum) 给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小. 说明:每次只能向下或者向右移动一步. 示例: 输入: [   [1,3,1], [1,5,1], [4,2,1] ] 输出: 7 解释: 因为路径 1→3→1→1→1 的总和最小. 找从左上角0,0到右下角的最短路径. DP:我们每个点(x,y)都可以表示为dp[x][y] = max( grid…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44371    Accepted Submission(s): 16084 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 330535    Accepted Submission(s): 78678 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…