m的范围没给，很坑爹

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12747    Accepted Submission(s): 4202

Problem DescriptionNow 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. ^_^ InputEach 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. OutputOutput the maximal summation described above in one line. Sample Input1 3 1 2 3
2 6 -1 4 -2 3 -2 3 Sample Output6
8

Hint

Huge input, scanf and dynamic programming is recommended.

 AuthorJGShining（极光炫影） 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <sstream>
#include <iostream>
using namespace std;
#define INF 0x3fffffff
typedef __int64 LL;
#define N 1000100LL dp[N][];
LL g[N];int main()
{
    //freopen("//home//chen//Desktop//ACM//in.text","r",stdin);
    //freopen("//home//chen//Desktop//ACM//out.text","w",stdout);
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(int i=;i<=n;i++)
        {
            g[i]=-;
            dp[i][]=dp[i][]=-;
        }
        int a=,b=;
        for(int i=;i<=n;i++)
        {
            swap(a,b);
            int tmp;
            scanf("%d",&tmp);
            for(int j=m;j>=;j--)
            {
                if(j>i) continue;
                dp[j][a]=max(dp[j][b],g[j-])+tmp;
                if(dp[j][a] > g[j]) g[j] = dp[j][a];
            }
        }
        printf("%I64d\n",g[m]);
    }
    return ;
}