给定一个浮点数数组，任意取出数组里若干个连续的数相乘，请找出其乘积最大的子 数组。要求用动态规划求出最大的连续乘积子数组，同时输出所求子数组的开始位置 和结束位置，请用C语言实现可运行的代码

以下是C语言实现可运行的代码：

#include <stdio.h>
#include <limits.h>

#define MAX_SIZE 100

int main() {
    int arr[MAX_SIZE];
    int n, i;
    int max_product = INT_MIN;
    int start_index, end_index, current_start_index;

    printf("Enter the size of the array: ");
    scanf("%d", &n);

    printf("Enter the elements of the array: ");
    for (i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }

    int max_ending_here = 1;
    int min_ending_here = 1;

    for (i = 0; i < n; i++) {
        if (arr[i] > 0) {
            max_ending_here *= arr[i];
            min_ending_here = (min_ending_here * arr[i] < 1) ? min_ending_here * arr[i] : 1;
        } else if (arr[i] == 0) {
            max_ending_here = 1;
            min_ending_here = 1;
            current_start_index = i + 1;
        } else {
            int temp = max_ending_here;
            max_ending_here = (min_ending_here * arr[i] > 1) ? min_ending_here * arr[i] : 1;
            min_ending_here = temp * arr[i];
        }

        if (max_ending_here > max_product) {
            max_product = max_ending_here;
            start_index = current_start_index;
            end_index = i;
        }
    }

    printf("The maximum product is %d\n", max_product);
    printf("The subarray with maximum product is: ");
    for (i = start_index; i <= end_index; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");

    return 0;
}

说明：

该算法使用动态规划的思想，通过记录当前位置前的最大和最小连续乘积，来计算当前位置的最大和最小连续乘积。同时记录下乘积最大的子数组的起始和结束位置，并输出。

时间复杂度为O(n)，空间复杂度为O(1)。