ACM学习感悟——HDU1204

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

这道题还是有点让我头疼的，主要是状态转移方程有点让人费解：dp[i][j]=MAX(dp[i][j-1] , dp[i-1][k])+data[j] 其中i

///////////////////////////////////////////////////////// 
//   HDU  2844     dp                                  //                        //
//  Created by 吴尔立 			                       //
//  Copyright (c) 2015年 吴尔立. All rights reserved.  //
/////////////////////////////////////////////////////////
#include 
#include 
#include 
#include 
#include            
#include 
#include 
#include 
#include 
#include 

				td[j-1]=temp;    //temp一直表示前一个状态的最大值。
				temp=max(temp,d[j]);
			} 
		}
		printf("%d\n",temp);
	}
}