A - Max Sum Plus Plus

最新推荐文章于 2020-10-20 23:32:14 发布

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本文探讨了一种使用动态规划解决最大子序列和问题的方法。给定一个整数序列，目标是找到特定数量的子序列对，使得这些子序列的和达到最大值。文章通过示例输入输出展示了算法的应用。

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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.

#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define MAXN 1000000

using namespace std;
int dp[MAXN+10];
int mmax[MAXN+10];
int a[MAXN+10];
int main()
{
    int n,m;
    int mmmax;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=0;
            mmax[i]=0;
        }
        dp[0]=0;
        mmax[0]=0;
        for(int j=1;j<=m;j++)
        {
            mmmax=-INF;
            for(int i=j;i<=n;i++)
            {
                dp[i]=max(dp[i-1]+a[i],mmax[i-1]+a[i]);//此时的mmax[i-1]对应j-1时候的值
                mmax[i-1]=mmmax;
                mmmax=max(dp[i],mmmax);
            }
        }
        printf("%d\n",mmmax);
    }
    return 0;
}