HDU

Max Sum Plus Plus

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 S ₁, S ₂, S₃, 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(i₁, 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. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S ₂, S ₃ ... S _n.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

68          

Hint

Huge input, scanf and dynamic programming is recommended.         

题意：

求最大m段子序列和

（可以直接用模板）

代码：

#include<stdio.h>  #include<algorithm>  #include<string.h>  #define N 1000000  #define INF 0x3f3f3f3fint a[N+5];  int now[N+5];  int pre[N+5];  using namespace std;  //最大m段和 （数组下标从一开始） int MaxSum(int m,int n,int *a){int maxx;memset(now,0,sizeof(now));      memset(pre,0,sizeof(pre));      for(int i=1;i<=m;i++)      {          maxx=-INF;          for(int j=i;j<=n;j++)          {              now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);              pre[j-1]=maxx;              if(now[j]>maxx)              {                  maxx=now[j];              }          }      }      return maxx;}int main()  {      int n,m,maxx,i,j;      while(scanf("%d%d",&m,&n)!=EOF)      {          for(i=1;i<=n;i++)          scanf("%d",&a[i]);                  printf("%d\n",MaxSum(m,n,a));      }      return 0;  }