Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41033    Accepted Submission(s): 14763 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

给定一个长度为n的区间:求m段连续子区间的和 最大值(其中m段子区间互不相交) 思路: dp[i][j]: 前j个元素i个连续区间最大值 (重要 a[j]必须在最后一个区间内) 转移方程:dp[i][j]=max (dp[i][j-1],dp[i-1][t]) + a[j]  ( dp[i-1][t] 是 max ( dp[i-1[k]  1<=k<=j-1 ) 第j个元素与第j-1个元素连在一起 --->dp[i][j-1]   第j个元素单独一个区间               --…

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…

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 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…

Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description:  System Crawler  (2015-09-05) Description Now I think you have got an AC in Ignatius.L's "Max Sum&…

Max Sum Plus Plus     HDU - 1024 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 a consecutiv…

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 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 Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23697    Accepted Submission(s): 8094 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…

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 faced with a more difficult problem. Given a consecutive number sequ…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21336    Accepted Submission(s): 7130 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

动态规划就是寻找最优解的过程 最重要的是找到关系式 hdu 1003 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目大意:求最大字序列和,其实就是分成 以0结尾的序列 以1结尾的序列 以2结尾的序列 ... 以n结尾的序列 所以以n结尾的序列的最大值就是以n-1结尾的序列的最大值+n的值 或最大值是n的值 关系式: d[i]=max(d[i-1]+a[i],a[i]) d[i]为以i结尾的序列和的最大值,a[i]为第i个数的值 #in…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1024     Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23844    Accepted Submission(s): 8143 Problem Description Now I think you hav…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题目大意:有多组输入,每组一行整数,开头两个数字m,n,接着有n个数字.要求在这n个数字上,m块数字的最大和.比如2 6 -1 4 -2 3 -2 3,就是(4 -2 3)和(3)这两块最大和为8. 解题思路:当成有m层,我们可以设置两个数组dp,mpre.dp[j]记录当前这一层包含a[j]时的最大值(包含a[j]),mpre[j]个记录上一层到第j-1个位置时的最大和(不一定包含a[j])…

Max Sum Plus Plus Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 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 mor…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35988    Accepted Submission(s): 12807 Problem Description Now I think you ha…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 刚开始的时候没看懂题目,以为一定要把那n个数字分成m对,然后求m对中和值最大的那对 但是不是,题目说的只是选出m对,所以有些数字是可以不用的. 那么就用 dp[i][j]表示前j个数,分成了i段,其中第a[j]个数必定包含在第i段之中的最大和值.就是a[j]必定选了而且在第i段之中. 至于为什么要这样设. 1.如果想得到ans,只需要扫描一次ans = max(ans, dp[m][m....n]),因…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意: 给定序列,给定m,求m个子段的最大和. 分析: 设dp[i][j]为以第j个元素结尾的i个子段的和. 对于每个元素有和前一个元素并在一起构成一个子段,和单独开启一个子段两种可能,状态转移方程 dp[i][j] = max(dp[i][j - 1], dp[i - 1][k]) + a[j] (k >= i - 1 && k <= j - 1) 时间复杂度O(m∗n2…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意:给定一个数组,求其分成m个不相交子段和的最大值. 这题有点问题其实m挺小的但题目并没有给出. dp[i][j]表示取第i 位的数共取了j段然后转移方程显然为 dp[i][j]=max(dp[i - 1][j]+a[j] , max(dp[j - 1][j - 1]~dp[i - 1][j - 1]))(大致意思是取第i位要么i-1位取了j个那么a[j]刚好能与i-1拼成一段,或者j -…

题目传送门//res tp hdu 数据范围1e6,若是开二维会爆 考虑用滚动数组优化 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define rep(i,a,b) for(int i=(a);i<=(b);++i) #define per(i,a,b) for(int i = (a);i>=(b);--i) #define fo(i,a,…

题意是要在一段数列中求 m 段互不重合的子数列的最大和. 动态规划,用数组 num[ ] 存储所给数列,建二维数组 dp[ ][ ] , dp[ i ][ j ] 表示当选择了第 j 个数字( num [ j ] )时,前 j 个数字被分成 i 组的所得最大和. 那么这个最大和等于 max{ ( 前 i - 1 组的和 + 第 i 组(含num[ j - 1]) + num[ j ] ),( 前 i - 1 组的和 + num[ j ] ) }: 即 dp[ i ][ j ] = max( dp…

子串和再续 时间限制:1000 ms  |  内存限制:65535 KB 难度:4 描述 给你一个序列 S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). 我们定义 sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).现在给你一个 m(8>m>0&&m<n)你的任务是计算 sum(i1, j1) + sum(i2, j2) + sum(i3,…

注:摘的老师写的 最大m子段和问题 以-1 4 -2 3 -2 3为例最大子段和是:6最大2子段和是:4+(3-2+3)=8所以,最大子段和和最大m子段和不一样,不能用比如先求一个最大子段和,从序列中去掉已求子段,再求下一个最大子段和的方法,这种方法有点贪心的味道,但是不行.所以,还得用动态规划. 1.基本思路:  首先,定义数组num[n],dp[m][n].   num[n]用来存储n个整数组成的序列. dp[m][n]代表n个整数序列求m个子段和. 按照分阶段的思想,我们首先考虑最后一项,…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25639    Accepted Submission(s): 8884 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

题解是看的这里的: http://www.acmerblog.com/hdu-1024-Max-Sum-Plus-Plus-1276.html 当前这个状态是dp[i][j],i 表示当前的段,j表示前j个数组成了当前的这i个段的最大值,而且a[j]在最后一个段中 状态dp[i][j]可以从dp[i][j-1]转移过来,表示第j个数字正好可以和 i 个段的前j-1个数字相加的和是当前所在状态中最大的 状态dp[i][j]可以从dp[i-1][j-1]转移过来,表示第 j 个数字正好可以成为第 i…

题意:最大和连续子序列的增强版,要求从一序列中取出若干段,这些段之间不能交叉,使得和最大并输出. 分析:用dp[i][j]表示前j个数取出i段得到的最大值,那么状态转移方程为dp[i][j]=max(dp[i][j-1]+a[j],dp[i-1][k]+a[j])  i-1<=k<=j-1 这个状态转移方程表达了两种不同的选择:第一个就是第j个连在第j-1个所在的段的后面,第二个就是第j个为新的一段的第一个数字. 由于数字的个数比较大,而题目中给定的m未知,怕超内存,所以要想办法开设一维数组来…