给定一个整数数组：（-2,1,-3,4,-1,2,1,-5,4）用分治法找出总和最大的连续数列，写出核心函数并返回总和。c++

这道题可以使用分治法来解决，将数组分成左半部分、右半部分和跨越中心的部分三个部分。其中，左半部分和右半部分可以分别递归求解，而跨越中心的部分则需要从中心向两边扩展来求解。

下面是核心函数的实现：

int maxSubArraySum(int arr[], int left, int right) {
    if (left == right) {
        return arr[left];
    }
    int mid = (left + right) / 2;
    int leftMaxSum = maxSubArraySum(arr, left, mid);
    int rightMaxSum = maxSubArraySum(arr, mid + 1, right);

    int leftCrossSum = INT_MIN, rightCrossSum = INT_MIN;
    int sum = 0;
    for (int i = mid; i >= left; i--) {
        sum += arr[i];
        if (sum > leftCrossSum) {
            leftCrossSum = sum;
        }
    }
    sum = 0;
    for (int i = mid + 1; i <= right; i++) {
        sum += arr[i];
        if (sum > rightCrossSum) {
            rightCrossSum = sum;
        }
    }
    int crossSum = leftCrossSum + rightCrossSum;

    return max(max(leftMaxSum, rightMaxSum), crossSum);
}

其中，`arr`为输入的整数数组，`left`和`right`为数组的左右边界。函数返回数组中的最大连续子序列和。

完整的代码如下：

#include <iostream>
#include <climits>

using namespace std;

int maxSubArraySum(int arr[], int left, int right) {
    if (left == right) {
        return arr[left];
    }
    int mid = (left + right) / 2;
    int leftMaxSum = maxSubArraySum(arr, left, mid);
    int rightMaxSum = maxSubArraySum(arr, mid + 1, right);

    int leftCrossSum = INT_MIN, rightCrossSum = INT_MIN;
    int sum = 0;
    for (int i = mid; i >= left; i--) {
        sum += arr[i];
        if (sum > leftCrossSum) {
            leftCrossSum = sum;
        }
    }
    sum = 0;
    for (int i = mid + 1; i <= right; i++) {
        sum += arr[i];
        if (sum > rightCrossSum) {
            rightCrossSum = sum;
        }
    }
    int crossSum = leftCrossSum + rightCrossSum;

    return max(max(leftMaxSum, rightMaxSum), crossSum);
}

int main() {
    int arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxSum = maxSubArraySum(arr, 0, n - 1);
    cout << "Max subarray sum is " << maxSum << endl;

    return 0;
}