Hdoj Max Sum plus plus "DP"

Max Sum plus Plus
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 18861 Accepted Submission (s): 6205

Problem Description

Now I think you has got an AC in IGNATIUS.L ' s "Max Sum" problem. To is a brave acmer, we always challenge ourselves to more difficult problems. Now you is faced with a more difficult problem.

Given a consecutive number sequence S1, S2, S3, S4 ... Sx... Sn (1≤x≤n≤1,000,000, -32768≤sx≤32767). We define a function sum (i, j) = Si + ... + Sj (1≤i≤j≤n).

Now given a integer m (M > 0), your task is to find m pairs of I and J which make sum (I1, J1) + sum (i2, J2) + sum (i3, J3) + ... + sum (IM, JM) maximal (IX≤IY≤JX or IX≤JY≤JX is not allowed).

But I ' m lazy, I don't want to the write a Special-judge module, so you don ' t has to output m pairs of I and j, just output th e maximal summation of sum (ix, JX) (1≤x≤m) instead. ^_^

Input

Each test case would begin with a integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1) 2 3
2 6-1 4-2) 3-2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

Author

Jgshining (Aurora Dazzle)
Test instructions: Find M continuous subsequence in the given n number, so that the sum of each sub-sequence is maximal.
Problem Solving Links
Code (not too understand, when to understand the time to fill in a detailed explanation):

#include <cstdio>#include <iostream>#include <cstring>Const intM =1e6+5;#define LL __int64using namespace STD; LL Pre[m], num[m];inlinell Max (ll A, ll B) {returnA > B? A:B;}voidDP (intMintN) {LL temp;//temp is dp[i][j]     for(inti =1; I <= m; + + i) {temp =0; for(intj =1; J <= I;        + + j) Temp + = Num[j]; Pre[n] = temp; for(intj = i+1; J <= N; + + j) {temp = max (temp, pre[j-1]) +num[j];//pre[j-1] is Max (Dp[i-1][t]) (i-1 <= T <= i-1);pre[j-1] = Pre[n];        Pre[n] = max (temp, pre[n]); }    }}intMain () {intN, M; while(scanf("%d%d", &m, &n) = =2){ for(inti =1; I <= N; + + i) {scanf("%i64d", num+i); Pre[i] =0; } DP (M, n);printf("%i64d\n", Pre[n]); }return 0;}

Hdoj Max Sum plus plus "DP"