hdu 1024 最大M字段和

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 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 an 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 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input
Each test case will begin with two 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

题意：

给出n个数，求其m个子序列和的最大值。

题解：

本人是没思路的。参考大佬图解:
http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html

代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;LL dp[2][1000001];LL a[1000001];LL res;int n,m;int main(){    while(scanf("%d%d",&m,&n)!=EOF)    {        int t=1;        res=0;        int i,j,k;        for(i=1;i<=n;i++)        {            dp[0][i]=dp[1][i]=0;            scanf("%lld",&a[i]);        }        for(i=1;i<=m;i++)        {            dp[t][i]=dp[1-t][i-1]+a[i];            LL MAX = dp[1-t][i-1];            for(j=i+1;j<=n-m+i;j++)            {                MAX = max(MAX,dp[1-t][j-1]);                dp[t][j]=max(dp[t][j-1],MAX)+a[j];            }            t=1-t;        }        t=1-t;        res=-(1<<30);        for(j=m;j<=n;j++)        {           res=max(res,dp[t][j]);        }        cout<<res<<endl;    }    return 0;}