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, 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. ^_^

Each 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.

Output the maximal summation described above in one line.

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

68   
       
         
     Hint    
    Huge input, scanf and dynamic programming is recommended.

#include
   
      #include
    
     #include
     
        using namespace std;int dp[1000010];int pre[1000010];int a[1000010];int main(){	int m, n;	int maxx;	while (~scanf("%d%d", &m, &n))	{		for (int i = 1; i <= n; i++)			scanf("%d", &a[i]);		memset(pre, 0, sizeof(pre));		for (int i = 1; i <= m; i++)		{			maxx = -1 << 28;			for (int j = i; j <= n; j++)			{				dp[j] = max(dp[j - 1], pre[j - 1]) + a[j];				pre[j - 1] = maxx;				if (maxx < dp[j])					maxx = dp[j];			}		}		printf("%d\n", maxx);	}	return 0;}