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 ... 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, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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. ^_^

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

Output the maximal summation described above in one line.

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

6
8


http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html
自己看吧。

#include <stdio.h>
#include <string.h>
#define INF 0x7fffffff
#define MAX 1000100

int dp[MAX] , num[MAX] , preMax[MAX];

int max(int a , int b)
{
	return a>b?a:b ;
}

int main()
{
	int m , n;
	while(~scanf("%d%d",&m,&n))
	{
		for(int i = 1 ; i <= n ; ++i)
		{
			scanf("%d",&num[i]) ;
			preMax[i] = 0 ;
			dp[i] = 0 ;
		}
		dp[0] = 0 ;
		preMax[0] = 0 ;
		int mm ;
		for(int i = 1 ; i <= m ; ++i)
		{
			mm = -INF;
			for(int j = i ; j <= n ; ++j)
			{
				dp[j] = max(dp[j-1]+num[j],preMax[j-1]+num[j]) ;
				preMax[j-1] = mm ;
				mm = max(dp[j],mm) ;
			}
		}
		printf("%d\n",mm) ;
	}
	return 0 ;
}