计数子阵列的总和在[L，R]范围内

Posted 2021-05-06

我正在解决竞争性编程问题，它被描述为：

给定n <10 ^ 5整数a1，a2，a3，...，an和L，R。有多少个子阵列，使得其元素之和在[L，R]范围内。

例：

输入：

n = 4, L = 2, R = 4
1 2 3 4

产量：4

(4 = 4, 3 = 1 + 2 = 3, 2 = 2)

我有一个解决方案是暴力，但是O（n ^ 2）太慢了。我应该使用哪些数据结构/算法来有效地解决这个问题？

答案

计算前缀和（p [0] = 0，p [1] = a1，p [2] = a1 + a2，...，p [n] =所有数字的总和）。

对于固定前缀sum p [i]，您需要找到j小于i且p [i] - R <= p [j] <= p [i]的此类前缀和p [j]的数量 - L.可以使用treap或其他平衡二叉搜索树在O（log n）中执行此操作。

伪代码：

treap.add(0)
sum = 0
ans = 0
for i from 1 to n:
    sum += a[i]
    left, right = treap.split(sum - R)
    middle, right = right.split(sum - L)
    ans += middle.size()
    merge left, middle and right together
    treap.add(sum)

另一答案

如果数组只包含正数，我们可以在线性时间内完成。

首先从左到右构建一个前缀为sum的数组。

1. Fix three pointers, X, Y and Z and initialize them with 0
2. At every step increase X by 1
3. While sum of numbers between X and Y are greater than R keep increasing Y
4. While sum of numbers between X and Z are greater than or equal to L, keep increasing Z
5. If valid Y and Z are found, add Z - Y + 1 to result.
6. If X is less than length of the array, Go to step 2.