HDU1003MAX SUM (动态规划求最大子序列的和)

【HDU1003MAX SUM (动态规划求最大子序列的和)】的更多相关文章

HDU1003MAX SUM (动态规划求最大子序列的和)

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…

HDU 1003 Max Sum【动态规划求最大子序列和详解】

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…

HDU 1003 Max Sum * 最长递增子序列（求序列累加最大值）

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 237978 Accepted Submission(s): 56166 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

HDU 1087 简单dp，求递增子序列使和最大

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384 Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

HDU1003MAX SUM

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…

ACM学习历程—HDU1003 Max Sum（dp && 最大子序列和）

Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the…

动态规划：HDU1003-Max Sum（最大子序列和）

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…

HDU 1081 To the Max 最大子矩阵（动态规划求最大连续子序列和）

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…

动态规划求最长公共子序列（Longest Common Subsequence, LCS）

1. 问题描述子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与母串保持一致,我们将其称为公共子序列.最长公共子序列(Longest Common Subsequence, LCS),顾名思义,是指在所有的子序列中最长的那一个.子串是要求更严格的一种子序列,要求在母串中连续地出现.在上述例子的中,最长公共子序列为blog(cnblogs, belong),最长公…

动态规划求一个序列的最长回文子序列（Longest Palindromic Substring ）

1.问题描述给定一个字符串(序列),求该序列的最长的回文子序列. 2.分析需要理解的几个概念: ---回文 ---子序列 ---子串 http://www.cnblogs.com/LCCRNblog/p/4321398.html这一篇文章描述了利用动态规划求解两个序列的最长公共子序列(Longest Common Sequence). 假设LCS(X,Y)表示序列X,Y的最长公共子序列,LPS(X)表示X的最长回文子序列: 在设序列X1为X的装置序列(逆序),比如X=“123”,X1=“32…

HDU 1024 Max Sum Plus Plus【动态规划求最大M子段和详解】

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…

HDU 1243 反恐训练营（动态规划求最长公共子序列）

反恐训练营 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3040 Accepted Submission(s): 693 Problem Description 当今国际反恐形势非常严峻,特别是美国"9.11事件"以后,国际恐怖势力更是有恃无恐,制造了多起骇人听闻的恐怖事件.基于此,各国都十分操心恐怖势力会对本国社会…

Max Sum Plus Plus---hdu1024（动态规划求M段的最大和）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意就是有n个数分成m段,求最大的和: dp[i][j]表示把 j 个数分成 i 段,选择第 j 个数的结果,而并不是当前的最优解, 那么要考虑的是第i个数的数是自己成一段还是和前面的成一段所以dp[i][j]=max(dp[i][j-1]+a[j], Max+a[j]); 其中Max为前 j-1 个数字分成 i-1 段中的最大值: 由于题中n为100w,m不知道是多少,所以开二维数组可能不…

Coincidence （动态规划求最长公共子序列）（王道）

题目描述: Find a longest common subsequence of two strings. 输入: First and second line of each input case contain two strings of lowercase character a…z. There are no spaces before, inside or after the strings. Lengths of strings do not exceed 100. 输出: Fo…

动态规划: HDU1003Max Sum

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…

UVa 108 - Maximum Sum（最大连续子序列）

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…

Leetcode 494 Target Sum 动态规划背包+滚动数据

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…

hdu1003 Max Sum【最大连续子序列之和】

题目链接:https://vjudge.net/problem/HDU-1003 题目大意:给出一段序列,求出最大连续子序列之和,以及给出这段子序列的起点和终点. 解题思路:最长连续子序列之和问题其实有很多种求解方式,这里是用时间复杂度为O(n)的动态规划来求解. 思路很清晰,用dp数组来表示前i项的最大连续子序列之和,如果dp[i-1]>=0的话,则dp[i]加上dp[i-1]能够使dp[i]增大:若dp[i-1]<0的话,则重新以dp[i]为起点,起点更新. #include <cs…

Max Sum(动态规划)

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…

HDU 1003 Max Sum (动态规划最大区间和）

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…

[leetcode]64Minimum Path Sum 动态规划

/** * 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. Note: You can only move either down or right at any point in time. */ /* * 动态规划题目,思路是把每一个点的最…

HDU 5791 Two（LCS求公共子序列个数）

http://acm.split.hdu.edu.cn/showproblem.php?pid=5791 题意: 给出两个序列,求这两个序列的公共子序列的总个数. 思路: 和LCS差不多,dp[i][j]表示第一个的前i个和第二个的前j个所包含的公共子序列的个数. 首先考虑a[i]≠b[j]的情况,此时应该容易推得dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]. 那么当a[i]=b[j]时,i和j这两个字符可以单独组成一个公共序列,然后前面dp[i-1][…

UVa 10891 - Game of Sum 动态规划，博弈难度: 0

题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1832 题意 n个数据,A,B两个玩家轮流从两端取1到多个数字,最终数字加和为分数,分数最大的获胜,A为先手,二者都很聪明,求A分数-B分数. 思路如刘书, 区间DP,明显可以用a[i][j]记录区间[i,j)的先手最大分数. 感想 1. 要像刘一样,在能达到要求之后进一步思考如何减…

动态规划--求最大连续子数组的和（Python实现）&求解最大连续乘积字串（Python实现）

def MaxSum(self,array,n): sum=array[0] result=array[0] for i in range(0,n): if sum<0: sum=a[i] else: sum=sum+a[i] start1=i if sum>result: result=sum end=i start=start1 print result,start,end 上述采用的是动态规划思想:假设sum[i]表示以第i个元素结尾的最大连续字串,那么sum[i]=max{sum[i-…

PAT甲题题解-1007. Maximum Subsequence Sum (25)-求最大子区间和

题意:给出n个数,求最大连续的子区间和,并且输出该区间的第一个和最后一个数. 如果所有数都小于0,那么则输出0,第一个数和最后一个数. 看数据k的范围,就知道肯定不能两层for循环来求区间和,O(n^2)的复杂度肯定超时所以这里肯定要求一遍for循环就能知道结果定义区间l和r,sum为目前[l,r]之间的和一开始l=r=0,sum=a[0]接下来,对于当前第i个数字a[i]如果sum+a[i]>a[i],那么就将a[i]一起加入到区间中去.如果sum+a[i]<a[i],那么还不如不加,直接重…

HDU-1003 Max Sum(动态规划，最长字段和问题)

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…