A

A - Max Sum Plus Plus

 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 sequence S ₁, S ₂, S ₃, S ₄ ... S _x, ... S _n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S _x ≤ 32767). We define a function sum(i, j) = S _i + ... + S _j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + sum(i _m, j _m) maximal (i _x ≤ i _y ≤ j _x or i _x ≤ j _y ≤ j _x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i _x, j _x)(1 ≤ x ≤ m) instead. ^_^

InputEach test case will begin with two integers m and n, followed by n integers S ₁, S ₂, S ₃ ... S _n.
Process to the end of file.
OutputOutput 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.
题意：给你n个数，让你从中取出m个子段使其和最大
题解：dp，dp[i][j]表示到a[j]包括a[j]从中去i段的最大值
所以dp[i][j]就分为两种情况：a[j]取或者不取
     dp[i%2][k]=max(dp[i%2][k-1],w[k]);
用w[i]记录一定取的情况，又分为两种情况：

   1、a[k]作为第i段

   2、a[k]加到之前的最大段那里
        w[k]=max(dp[(i-1)%2][k-1],w[k-1])+sum[k]-sum[k-1];

初值：  dp[0][i]               
    if(i==k）dp[i%2][k]=w[k]=sum[k];

具体看代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define INF 0x3f3f3f3f
#define N 1000005
int n,m;
int w[N];
int dp[2][N];
int sum[N];
int a[N];
int main()
{
     ios_base::sync_with_stdio(0); cin.tie(0);
    while(cin>>m>>n){
    sum[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
        dp[0][i]=0;
    }
     for(int i=1;i<=m;i++)
       {
           for(int k=i;k<=n;k++)
           {
               if(i==k)
               dp[i%2][k]=w[k]=sum[k];//从k个数中取k段的最大值是前k个数的和
               else
               {
                   w[k]=max(dp[(i-1)%2][k-1],w[k-1])+sum[k]-sum[k-1];//这是一定要取的情况，分为两种：1、a[k]作为第i段，2、a[k]加到之前的最大段那里
                   dp[i%2][k]=max(dp[i%2][k-1],w[k]);//a[k]取或者不取
               }
           }
       }

    cout<<dp[m%2][n]<<endl;
    }
    return 0;

}

参考博客：http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html

        https://blog.csdn.net/lishuhuakai/article/details/8067474