解法一：

双指针，没必要开数组直接边输边算，max至少要2个数，补一个数时的特判

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define endl '\n'

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	long long n, a, cnt = 0, maxsum = 0;
	cin >> n;
	cin >> a; n--;
	cnt += a;
	maxsum = cnt;
	while (n--) {
		cin >> a;
		cnt += a;
		maxsum = max(maxsum, cnt);
		if (cnt <= 0) cnt = 0;
	}
	cout << maxsum << endl;
	return 0;
}

解法二：

dp[i]:以vec[i]为段尾的最大子段和

状态转移方程：

base condition：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define endl '\n'

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	vector<long long> vec(n, 0);
	vector<long long> dp(n, 0);
	for (int i = 0; i < n; i++) cin >> vec[i];
	dp[0] = vec[0];
	for (int i = 1; i < n; i++) {
		dp[i] = max(dp[i - 1] + vec[i], vec[i]);
	}
	cout << *max_element(dp.begin(), dp.end()) << endl;
	return 0;
}