1007. Maximum Subsequence Sum (25)

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本文探讨了最大子序列和问题的两种解决方法，并通过示例详细解释了如何找到具有最大元素和的连续子序列及其边界值。文章还讨论了特殊情况下的处理方式。

1007. Maximum Subsequence Sum (25)

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

#include <iostream>

using namespace std;
int Array[10005];
int large_sum = -1, first, last;
// 常规的做法:二次遍历，以每个点为起点，找子串和最大时的终点。
void solve1(int K){
  for(int i = 0; i < K; i++){
    if(Array[i] >= 0){
      int sum = 0;
      for(int j = i; j < K; j++){
        sum += Array[j];
        if(sum > large_sum){
          large_sum = sum;
          first = i;
          last = j;
        }
      }
    }
  }
}
//第二种解法：保存一个最大字段和以及一个当前子段和, 如果当前字段和大于当前最大字段和, 那么更新这个最大字段和, 如果当前字段和为负数的时候, 直接把当前字段和甚设置成0,注意记录边界first 与 last 的值
void solve2(int K){
  int sum = 0;
  int tem_first = 0;
  for(int i = 0; i < K; i++){
    sum += Array[i];
    if(sum > large_sum){
      large_sum = sum;
      last = i;
      first = tem_first;
    }
    if(sum < 0){
      sum = 0;
      tem_first = i + 1;
    }
  }
}
int main(){
  int K;
  bool flag = false;
  cin >> K;
  for(int i = 0; i < K; i++){
    cin >> Array[i];
    if(Array[i] >= 0)
      flag = true;
  }
  if(flag){
    solve2(K);
    cout << large_sum << " " << Array[first] << " " << Array[last] << endl;
  }else{
    cout << 0 << " " << Array[0] << " " << Array[K-1] << endl;
  }
  return 0;
}