PTA 1007 Maximum Subsequence Sum (25 分)（详细翻译）

悠悠 2022-09-14 11:13 158阅读 0赞

# 背景： #

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.  
**给定一个有K个整数的序列 \{ N 1 , N 2 , …, N K \}。 一个连续的子序列定义为 \{N i , N i+1 , …, N j \} 其中 1 ≤ i ≤ j ≤ K。 最大子序列是其元素之和最大的连续子序列。 例如，给定序列 \{ -2, 11, -4, 13, -5, -2 \}，其最大子序列为 \{ 11, -4, 13 \}，最大和为 20。**

# 输入格式： #

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.  
**每个输入文件包含一个测试用例。 每个测试用例占据两行。 第一行包含一个正整数 K (≤10000)。 第二行包含 K 个数字，用空格分隔。**

# 输出格式： #

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.  
**对于每个测试用例，在一行中输出最大的和，以及最大子序列的第一个和最后一个数字。 数字必须由一个空格分隔，但行尾不得有多余空格。 如果最大子序列不唯一，则输出具有最小索引 i 和 j 的子序列（如示例案例所示）。 如果所有 K 个数字都是负数，那么它的最大和被定义为 0，你应该输出整个序列的第一个和最后一个数字。**

# 输入样例： #

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

# 输出样例： #

10 1 4

# 思路： #

本题是`求最大子序列和`问题。使用经典的算法，再稍加修改即可。

## C++实现代码： ##

#include <iostream>
    #include<vector>
    using namespace std;
    int main()
    { 
        int K;
        cin >> K;
        vector<int> num;
        while (K--)
        { 
            int i;
            cin >> i;
            num.push_back(i);
        }
        int maxSum = -1, thisSum = 0, beg = 0, end = 0, spe = 0;
        for (decltype(num.size()) i = 0; i < num.size(); ++i)
        { 
            thisSum += num[i];
            if (thisSum == num[i])
                spe = i;
            if (thisSum > maxSum)
            { 
                maxSum = thisSum;
                beg = spe;
                end = i;
            }
            else if (thisSum < 0)
                thisSum = 0;
        }
        if (maxSum < 0)
        { 
            maxSum = 0;
            beg = 0;
            end = num.size() - 1;
        }
        cout << maxSum << ' ' << num[beg] << ' ' << num[end] << endl;
        return 0;
    }