#include using namespace std; const int INF = 0x3f3f3f3f; const int N = 200010; int n; int a[N]; //分治函数 int rec(int l, int r) { //递归的出口 if (l == r) return a[l]; // l=r时,直接返回该位置的值 int mid = (l + r) >> 1; //中间点 // [l..mid]区间内包含mid的最大后缀和 int sum = 0, lmax = -INF, rmax = -INF; for (int i = mid; i >= l; i--) sum += a[i], lmax = max(lmax, sum); // [mid+1..r]区间内包含(mid+1)的最大前缀和 sum = 0; for (int i = mid + 1; i <= r; i++) sum += a[i], rmax = max(rmax, sum); //三种可能取最大值 return max(max(rec(l, mid), rec(mid + 1, r)), lmax + rmax); } int main() { cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; printf("%d", rec(1, n)); //在范围1~n之间求最大子段和 return 0; }