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;
}