浙大MOOC数据结构 编程题 01-复杂度2 Maximum Subsequence Sum

题目描述：

Given a sequence of K integers {N1, N2 , ..., NK}. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj} 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

Sample Output:

10 1 4

解决思路：

采用上一题的解法四 在线处理 思路，并记录 res 的左右坐标，最后直接输出。
值得一提的是，本次是先判断 st 是否为最大子序列，再去将 st < 0 时的 st 归零（重新记录连续子序列）
好处有三：
1、可以使 res 的值为连续最大子序列
2、若子序列全为负数，则 res 为子序列最大值（依旧 < 0，便于判断特殊情况）
3、若子序列中最大值为0，则 res 值为0 （= 0时，便于判断临界情况）
PS：上一题中先归零再判断，会使 res 值最小为0

#include <iostream>

using namespace std;

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

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int st = 0, res = -1;
    int st_l = 0, res_l = 0, res_r = 0;
    for (int i = 0; i < n; i++)
    {
        st += a[i];
        if (st > res)
        {
            res = st;
            res_l = st_l;
            res_r = i;
        }
        if (st < 0)
        {
            st = 0;
            st_l = i + 1;
        }
    }
    if (res < 0) cout << "0 " << a[0] << ' ' << a[n - 1] << endl;
    else cout << res << ' ' << a[res_l] << ' ' << a[res_r] << endl;
    return 0;
}