01-复杂度2 Maximum Subsequence Sum

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该程序解决了一个经典的计算机科学问题，即找出一个整数序列中具有最大和的连续子序列。输入包含一个正整数K，然后是K个整数。通过遍历序列，维护当前子序列的和，并更新最大子序列和及其起始位置。如果所有数字都是负数，则最大和为0，输出整个序列的首尾元素。

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.

#include <iostream>

using namespace std;

int main()
{
    //输入数据
    int K;
    cin >> K;
    int data[K];
    int i = 0;
    while(i < K){
        cin >> data[i];
        i++;
    }
    //最大子列和问题  返回对应的数组下标
    int cursum = 0;   //当前和
    int ret = 0;      //最大子列和
    int templ = 0;
    int l = 0, r = 0;
    int len = 0;
    for(int i = 0; i < K; i++){
        cursum += data[i];
        if(cursum < 0){
            cursum = 0;
            templ = i + 1;
        }
        if(data[i] < 0) len++;
        if(ret < cursum || (ret == 0 && data[i] == 0)){
            ret = cursum;
            l = templ;
            r = i;
        }
    }
    if(len == K){
        l = 0;
        r = K - 1;
    }
    
    cout << ret << " " << data[l] << " " << data[r] << endl;
    return 0;
}