#include 
    
     using namespace std;int Find_Max_Crossing_SubAr(int A[], int low, int mid, int high,int *max_left,int *max_right){   int left_sum = -10000000;   int sum = 0;   for (int i = mid; i >= low; i --)   {      sum += A[i];      if (sum >left_sum)      {         left_sum = sum;         *max_left=i;      }   }   int right_sum = -100000000;   sum = 0;   for (int j = mid + 1; j <= high; j ++)   {      sum += A[j];      if (sum > right_sum)      {         right_sum = sum;         *max_right=j;      }   }   return left_sum + right_sum;}int Find_Maximum_SubAr(int A[], int low, int high,int *max_left,int *max_right){   int left_sum, right_sum, cross_sum;   if (high == low)   {      *max_left=low;      *max_right=high;      return A[low];   }   else   {      int mid = (low + high) / 2;      left_sum = Find_Maximum_SubAr(A, low, mid,max_left,max_right);      right_sum = Find_Maximum_SubAr(A, mid + 1, high,max_left,max_right);      cross_sum = Find_Max_Crossing_SubAr(A, low, mid, high,max_left,max_right);      if (left_sum >= right_sum && left_sum >= cross_sum)      {         return left_sum;      }      else if (right_sum >= left_sum && right_sum >= cross_sum)      {         return right_sum;      }      else      {         return cross_sum;      }   }}int main(){    int A[100];    int n;    cout<<"Input the number of numbers:";    cin>>n;    for (int i = 0; i < n; i ++)    {       cin>>A[i];    }    int s,l;    cout<<"The max sum of the subarray is："<
    

该算法复杂度O（n*logn）；

主要思想：最大子数组区间必定存在于数组中点左方（不包括中点处）或跨越中点或数组中点右方（不包括中点）的区域，再使用递归便可求出。