HDU 1594 find the max

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数序问题. 题意是说一个数列 a1,a2,--ai,--an; x=i , y = ai:找两个点斜率绝对值.!最大. 第一次没找绝对值,--认真读题. .. x 每次加1 . 仅仅须要找相邻的 ai 是否是最大就好了. #include<cstdio> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<map> #in…

hdu 1754 线段树(Max+单点修改)

I Hate It Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 57279 Accepted Submission(s): 22365 Problem Description 很多学校流行一种比较的习惯.老师们很喜欢询问,从某某到某某当中,分数最高的是多少. 这让很多学生很反感. 不管你喜不喜欢,现在需要你做的是,就是按照老师…

hdu 1081 To The Max（dp+化二维为一维）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1081 To The Max Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8839 Accepted Submission(s): 4281 Problem Description Given a two-dimensional ar…

dp - 最大子矩阵和 - HDU 1081 To The Max

To The Max Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=1081 Mean: 求N*N数字矩阵的最大子矩阵和. analyse: 乍看题目意思很简单,但对于刚开始学DP的新手来说也不是很简单. 这道题使用到的算法是:预处理+最大连续子串和如果会做最大连续子串和,那么理解这题就相对简单一些,若不知道最大连续子串和,建议先看一下这两题: http://acm.hdu.edu.cn/showproblem.php?pi…

HDU 1081 To The Max【dp，思维】

HDU 1081 题意:给定二维矩阵,求数组的子矩阵的元素和最大是多少. 题解:这个相当于求最大连续子序列和的加强版,把一维变成了二维. 先看看一维怎么办的: int getsum() { ; int ans=-1e9; ;i<=n;i++){ ) tot=; tot+=a[i]; if(tot>ans) ans=tot; } return ans; } 这种做法太棒了!短短几行,就能解决最大子序列和这个问题.其实这几行代码值得深思.而且这是个在线算法,输入数据及时能给出结果,感觉不能归于动归…

Hdu 1081 To The Max

To The Max Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7620 Accepted Submission(s): 3692 Problem Description Given a two-dimensional array of positive and negative integers, a sub-rectang…

URAL 1146 Maximum Sum & HDU 1081 To The Max (DP)

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 &…

HDU 1081 To The Max(动态规划)

题目链接 Problem Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in th…

hdu 1003,nefu 728 max sum

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 input contains an inte…

ACM HDU 1081 To The Max

To The Max Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10839 Accepted Submission(s): 5191 Problem Description Given a two-dimensional array of positive and negative integers, a sub-re…

hdu 1081 To The Max(二维压缩的最大连续序列)(最大矩阵和)

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

hdu 3415(单调队列) Max Sum of Max-K-sub-sequence

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=3415 大意是给出一个有n个数字的环状序列,让你求一个和最大的连续子序列.这个连续子序列的长度小于等于k. 由于是环状的数列,一般采取的措施是在数列的后面再复制一部分前面的数列;a[n]=a[0],a[n+1]=a[1]---- 用sum数组存储前i个数的和,那么对于j属于范围(i-k,i-1)来说,最大连续子序列就等于sum[i]减去在该范围内最小的sum[j] 那么问题就变成了求在每k个范围内最小的…

HDU 1081 To The Max （dp）

题目链接 Problem Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in th…

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…

HDU 1081 To The Max - 最大字段和

传送门题目大意: 求一个矩阵的最大子矩阵和. 题目分析: 刚开始考虑了一下dp方程的递推,但是不好转.简便的方法是预处理sum[i][j]表示第i行的前j个元素之和,之后\(n^3\)枚举子矩阵就可以了. code #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #…

HDU 5317 RGCDQ (数论素筛)

RGCDQ Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Submit Status Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with Ra…

HDU 1024 max sum plus

A - Max Sum Plus Plus Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

hdu 1003 MAX SUM 简单的dp，测试样例之间输出空行

测试样例之间输出空行,if(t>0) cout<<endl; 这样出最后一组测试样例之外,其它么每组测试样例之后都会输出一个空行. dp[i]表示以a[i]结尾的最大值,则:dp[i]=max(dp[i]+a[i],a[i]) 解释: 以a[i]结尾的最大值,要么是以a[i-1]为结尾的最大值+a[i],要么是a[i]自己本身,就是说,要么是连同之前的构成一个多项的字串,要么自己单独作为一个字串,不会有其他的可能了. 状态规划的对状态的要求是:当前状态只与之前的状态有关,而且不影响下一…

HDU 2993 MAX Average Problem（斜率优化）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2993 Problem Description Consider a simple sequence which only contains positive integers as a1, a2 ... an, and a number k. Define ave(i,j) as the average value of the sub sequence ai ... aj, i<=j. Let’s…

HDU 1024：Max Sum Plus Plus（DP）

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem 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 fac…

HDU 1024 Max Sum Plus Plus --- dp+滚动数组

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面方程: a[j]直接接在前面…

HDU 1003 Max Sum --- 经典DP

HDU 1003 相关链接 HDU 1231题解题目大意:给定序列个数n及n个数,求该序列的最大连续子序列的和,要求输出最大连续子序列的和以及子序列的首位位置解题思路:经典DP,可以定义dp[i]表示以a[i]为结尾的子序列的和的最大值,因而最大连续子序列及为dp数组中的最大值. 状态转移方程:dp[1] = a[1]; //以a[1]为结尾的子序列只有a[1]: i >= 2时, dp[i] = max( dp[i-1]+a[i], a[i] ); dp[i-1]+a[i…

HDU 1244 Max Sum Plus Plus Plus

虽然这道题看起来和 HDU 1024 Max Sum Plus Plus 看起来很像,可是感觉这道题比1024要简单一些前面WA了几次,因为我开始把dp[22][maxn]写成dp[maxn][22]了,Orz 看来数组越界不一定会导致程序崩溃,也有可能返回一个错误的结果 dp[i][j]表示前j个数构成前i段所得到的最大值状态转移方程: dp[i][j] = max{dp[i][j-1], dp[i-1][j-len[i]] + sum[j] - sum[j-len[i]]} 分别对应…

HDU 2993 MAX Average Problem（斜率优化DP）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2993 题目大意:给定一个长度为n(最长为10^5)的正整数序列,求出连续的最短为k的子序列平均值的最大值. Sample Input 10 6 6 4 2 10 3 8 5 9 4 1 Sample Output 6.50 分析:斜率优化DP,要认真看代码如下: # include<iostream> # include<cstdio> # include<cstring&…

数据结构：HDU 2993 MAX Average Problem

MAX Average Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7639 Accepted Submission(s): 1667 Problem Description Consider a simple sequence which only contains positive integers as a…

【动态规划】HDU 1081 & XMU 1031 To the Max

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1081 http://acm.xmu.edu.cn/JudgeOnline/problem.php?id=1031 题目大意: 给一个n*n(n<=100)的矩阵,求一个矩形覆盖的值最大是多少. 题目思路: [动态规划] 二维的最大字段和.先考虑一维的情况.f[i]=max(f[i-1]+a[i],a[i]) 只要之前的部分和大于零则一起取一定比只取当前位置的要优. 因此只要判断局部段的和是否大于零…