【PAT甲级 - C++题解】1007 Maximum Subsequence Sum

✍个人博客：https://blog.csdn.net/Newin2020?spm=1011.2415.3001.5343
📚专栏地址：PAT题解集合
📝原题地址：题目详情 - 1007 Maximum Subsequence Sum (pintia.cn)
🔑中文翻译：最大子序列和
📣专栏定位：为想考甲级PAT的小伙伴整理常考算法题解，祝大家都能取得满分！
❤️如果有收获的话，欢迎点赞👍收藏📁，您的支持就是我创作的最大动力💪

1007 Maximum Subsequence Sum

Given a sequence of K integers { N₁, N₂, …, N_K }. A continuous subsequence is defined to be { N_i, N_i+1, …, N_j } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21
1
2

Sample Output:

10 1 4
1

题意

最大子序列是指序列内各元素之和最大的连续子序列。现在你需要求出最大子序列的各元素之和，并且输出最大子序列的第一个元素和最后一个元素的值。

思路

状态表示： $f[i]$ 表示所有以 $i$ 为右端点的区间中区间和的最大值。

状态计算： $f[i]=max(w_i,f[i-1]+w_i)=w_i+max(f[i-1],0)$

我们在写代码的时候就可以用一个变量 f 来代替数组，表示当前区间和，然后用一个 res 来记录最大的区间和。

代码

#include
using namespace std;

const int N = 10010;
int w[N];
int n;

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)    cin >> w[i];

    //从前往后遍历，动态更新答案
    int res = -1, l, r;
    for (int i = 0, f = -1, start; i < n; i++)
    {
        //如果f小于0，则抛弃当前区间，从i开启新的区间
        if (f < 0) f = 0, start = i;
        f += w[i];
        if (res < f)   //如果当前连续区间和大于res，则更新
        {
            res = f;
            l = w[start], r = w[i];
        }
    }

    if (res < 0)   res = 0, l = w[0], r = w[n - 1];

    cout << res << " " << l << " " << r << endl;

    return 0;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32