Max Sum Plus Plus

Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24165    Accepted Submission(s): 8274
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 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. ^_^
 
Input
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
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


    
    
    
    
 
     
 
      Hint 
     
 Huge input, scanf and dynamic programming is recommended. 
    
    
    
    
     
   
   
   
   

 
Author
JGShining
n個の数の中からm個のセグメントを探し出してそれと最大にさせます
ACコード:

#include<iostream>
#include<cstdio>
#include<cmath> 
#include<cstring>
using namespace std;
#define M 1000001 
int dp[M],t[M],num[M],n,m;
int main()
{
    int i,j;
    int Max;
    while(~scanf("%d %d",&m,&n)){
        for(i=1;i<=n;i++)
        scanf("%d",&num[i]);
        memset(dp,0,sizeof(dp));
        memset(t,0,sizeof(t));
        for(i=1;i<=m;i++){
            Max=INT_MIN;
            for(j=i;j<=n;j++){
                dp[j]=max(dp[j-1],t[j-1])+num[j];
                t[j-1]=Max;
                Max=max(dp[j],Max);
            }
            t[j-1]=Max;
        }
        printf("%d
",Max);
    }
}