Geometric Mean Maximum Subarray

topic description

Find the subarray with length at least and maximum geometric mean from an Narray of positive numbers of length , and output its position and size. (The geometric mean of K numbers is the Kth root of the product of K numbers) If there are multiple sub-arrays whose geometric mean is the maximum value, then output the sub-array with the smallest length. If there are multiple subarrays with the same length whose geometric mean is the maximum value, then output the first subarray.numbersL

enter description

• The first line of input isN、L
  • NRepresented numberssize $1\le N\le 100000$
  • LIndicates the minimum length of a subarray $1\le L\le N$
• The following Nlines represent the number numbersin Neach line $10^{-9}\le numbers[i]\le 10^9$

output description

• Output the position (counting from 0) and size of the subarray, separated by a space.

Remark

• The use case guarantees that, except for the subarray whose geometric mean is the maximum value, the geometric mean of the other subarrays is at least 1 smaller than the maximum value $10^{-\; Multiples\; of10}$ .

Example

use case 1

• enter

3 2
2
2
3

• output

1 2

• illustrate
• There are 3 sub-arrays with length at least 2, which are{2, 2}、{2, 3}、{2, 2, 3}
• Among them {2, 3}, the geometric mean is the largest, so output its position 1 and length 2

use case 2

• enter

10 2

0.2
0.1
0.2
0.2
0.2
0.1
0.2
0.2
0.2
0.2

• output

2 2

• illustrate
• There are multiple subarrays of length at least 2 that have a geometric mean of 0.2, among which the shortest length is 2, and there are multiple.
• The first subarray of length 2 and geometric mean of 0.2 is a subarray composed of two 0.2 starting from the second number

code display

    private static void maxGeoSubArray(double[] numbers, int L) {
    
    
        // 首先构造前缀积数组
        int n = numbers.length;
        double[] mul = new double[n+1];
        // 初始化第一个元素 为 0
        mul[0] = 1.0;

        // 填充前缀积数组数据
        for(int i = 0;i < n;i++) {
    
    
            mul[i + 1] = mul[i] * numbers[i];
        }

        // 初始化左右边界变量
        int l = 0,r = 0;
        // 保存最大值
        double ans = 0.0;

        //  双重遍历，去找到 最大几何平均值
        for(int i = 0;i <= n - L;i++) {
    
    
            for(int j = i + L - 1;j < n;j++) {
    
    
                // 这个时候，计算区间  [i, j] 的几何平均值 j - i + 1
                double t = mul[j+1] / mul[i];
                int len = j - i + 1;
                //  开  len 次方根
                double avg = Math.pow(t, 1.0 / len);

                // 根据备注要求，几何平均值的大小至少要保证比最大值 小 10^{-10} 次方倍
                if(Math.abs(avg - ans) > Math.max(avg, ans) / Math.pow(10, 10) && avg >= ans) {
    
    
                    if(avg == ans) {
    
    
                        int local = r - l + 1;
                        if(len > local) {
    
    
                            r = j;
                            l = i;
                        }
                    } else {
    
    
                        ans = avg;
                        r = j;
                        l = i;
                    }
                }
            }
        }

        // 最终遍历最后然后输出结果   输出  位置 和 长度
        System.out.print(l + " ");
        System.out.print(r - l + 1);
    }