1007. Maximum Subsequence Sum (25)

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本文探讨了如何找出整数序列中拥有最大元素和的连续子序列，并详细介绍了实现这一功能的具体算法。针对不同的输入情况，如全部为负数或包含零的情况，文章提供了相应的解决方案。

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

Sample Output:

10 1 4

测试点5要小心,因为都为负数才输出0 第一位最后一位。而如果有一个数的0的话最大值是0，此时i和j也都是0

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int n, num, flag = 0;
    int thisnum = 0, maxnum = -1, tmpn = -1, minn = 0, maxn = 0;
    vector<int> a;
    cin >> n;
    for(int i=0; i<n; i++)
    {
        cin >> num;
        if(num >= 0)
            flag = 1;
        a.push_back(num);
    }
    for(int i=0; i<n; i++)
    {
        thisnum += a[i];
        if(thisnum < 0)
        {
            thisnum = 0;
            tmpn = i;
        }
        if(thisnum > maxnum)
        {
            maxnum = thisnum;
            maxn = a[i];
            minn = a[tmpn+1];
        }
    }
    if(maxnum==0)//测试点5
    {
        maxn = 0;
        minn = 0;
    }
    if(flag==0)
    {
        maxnum = 0;
        minn = a[0];
        maxn = a[n-1];
    }
    cout << maxnum << " " << minn << " " << maxn;
    return 0;
}