1007 Maximum Subsequence Sum （25 分）

Given a sequence of K integers { N​1​​, N​2​​, …, 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

Code

#include <iostream>
#include <vector>

using namespace std;

int main()
{
	int n;
	cin >> n;
	vector<int> v(n);
	int leftIndex = 0, rightIndex = n - 1, sum = -1, temp = 0, tempIndex = 0;
	for (int i =
 
 0; i < n; i++) {
		cin >> v[i];
		temp += v[i];
		if (temp < 0) {
			temp = 0;
			tempIndex = i + 1;
		}
		else if (temp > sum) {
			sum = temp;
			leftIndex = tempindex;
			rightIndex = i;
		}
	}
	if (sum < 0) sum = 0;
	cout << sum << " " << v[leftIndex] << " " << v[rightIndex] << endl;
	return 0;
}

思路

用vector v(n)儲存所有輸入的數，要輸出的最大字串和儲存在sum中，leftindex、rightindex分別儲存其左端點和右端點，temp為當前字串和，tempIndex為temp的左端點。從左到右掃描v(n)，temp為累加和，當掃描到第i個數時，如果temp += v[i]後，temp為負，那麼temp置為0，tempIndex置為i+1，因為temp+v[i+1] < v[i+1]；如果temp > sum那麼更新sum的值和左右端點。
本文程式碼參考https://blog.csdn.net/liuchuo/article/details/52144554感謝這位博主的分享！
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