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…
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 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…
[题解]最大 M 子段和 Max Sum Plus Plus [Hdu1024] [51nod1052] 传送门:最大 \(M\) 子段和 \(Max\) \(Sum\) \(Plus\) \(Plus\) \([Hdu1024]\) \([51nod1052]\) [题目描述] 给出一个长度为 \(N\) 的序列 ,将这 \(N\) 个数划分为互不相交的 \(M\) 个子段,并使得 \(M\) 个子段的和最大. [样例] 样例输入: 7 2 -2 11 -4 13 -5 6 -2 样例输出:…
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): 21336    Accepted Submission(s): 7130 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…
题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1024 (http://www.fjutacm.com/Problem.jsp?pid=1375) 题意:长度为n的序列里,m段不相关区间的最大和 思路:我们先要确定一个东西,就是状态,这里我用dp[i][j]表示前j个数在取a[j]情况下分i段的最大和: 那么我们为了找规律,可以先来一发Excel,就以样例为例子: 然后我们可以发现其实红圈里的8是状态dp[2][6](i=2, j=6),那么我们可…
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…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41885    Accepted Submission(s): 15095 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…