数据结构-最大子列和问题-实现返回最大子列和，最大子列的首尾数值

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本文介绍了一个寻找整数序列中最大连续子序列及其和的算法。该算法能够处理包含正负数的序列，输出最大子序列的起始和结束元素及总和。若所有数均为负，则返回整个序列的边界值及定义的最大和为0。

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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

程序：

#include<iostream>
using namespace std;

int getMaxSum2(int A[], int N);

int main()
{
    int N=0;
    cin >> N;
    int arr[N];
    int i = 0;
    char c;
    //以空格为一个数值的输入，回车确定
    cin >> arr[i++];
    while((c=getchar())!='\n')
    {
        cin >> arr[i++];
    }
    getMaxSum2(arr, N);
    return 0;
}

int getMaxSum2(int A[], int N)
{
    //ThisSum当前子列和；MaxSum最大子列和
    int ThisSum, MaxSum;
    int i;
    //当前子列的首尾下标
    int thisLeft, thisRight;
    //当前最大子列的首尾下标
    int maxLeft, maxRight;
    //true：表示当前子列全为负数
    bool allPo = true;

    //-1表示子列的首下标可变
    thisLeft = thisRight = -1;
    maxLeft = 0; maxRight = N-1;

    ThisSum = 0; MaxSum = -1;
    for(i=0; i<N; i++)
    {
        if(A[i] >= 0)
            allPo = false;
        ThisSum += A[i];
        if(ThisSum>=0)
        {
            if(thisLeft == -1)
            {
                thisLeft = i;

            }
            thisRight = i;
        }
        if(ThisSum > MaxSum)
        {
            MaxSum = ThisSum;
            maxLeft = thisLeft;
            maxRight = thisRight;
        }
        else if(ThisSum<0)
        {
            ThisSum = 0;
            thisLeft = -1;
            thisRight = -1;
        }

    }
    if(allPo == true)
    {
        maxLeft = 0;
        maxRight = N-1;
        MaxSum = 0;
    }
    printf("%d %d %d\n", MaxSum, A[maxLeft], A[maxRight]);
}