hdu 1024Max Sum Plus Plus

最新推荐文章于 2024-07-21 14:30:00 发布

wb_james

最新推荐文章于 2024-07-21 14:30:00 发布

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文章标签： output integer module function input im

本文链接：https://blog.youkuaiyun.com/nbu_james/article/details/4632271

本文探讨了一种复杂版本的最大子序列和问题，该问题要求在给定的整数序列中找到多个不重叠子序列，使得这些子序列的和达到最大值。文章提供了一个高效的解决方案，并通过样例输入输出展示了解决方案的有效性。

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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 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. ^_^

Input

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

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

题目是看懂了，但代码还没理解！！

#include<iostream>
using namespace std;
#define max(a,b) a>b?a:b
#define MAXN 1000000
int num[MAXN+50],now[MAXN+50],pre[MAXN+50];
int main()
{
 int m,n,i,j,k,ans,max_pre;
    while(cin>>m>>n)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);                    //数据太多所以用scanf 
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(i=1;i<=m;i++)
        {
   max_pre=-99999999;
            for(j=i;j<=n;j++)
            {
    
                now[j]=max(now[j-1]+num[j],pre[j-1]+num[j]);
                pre[j-1]=max_pre;
                if(now[j]>max_pre)
                    max_pre=now[j];
             //cout<<pre[j-1]<<'*'<<now[j-1]<<'*'<<now[j]<<'*'<<max_pre<<endl;
   }
   cout<<endl;
  }
        cout<<max_pre<<endl;
    }
 
    return 0;
}