2026-06-08：开销小于等于 K 的子数组数目。用go语言，给定整数数组 nums 和整数 k。 对数组中任意一个连续非空子数组 nums[l..r]，先找

package main

import (
    "fmt"
)

func countSubarrays(nums []int, k int64) (ans int64) {
    var minQ, maxQ []int
    left := 0
    for right, x := range nums {
        // 1. 入：元素进入窗口
        forlen(minQ) > 0 && x <= nums[minQ[len(minQ)-1]] {
            minQ = minQ[:len(minQ)-1]
        }
        minQ = append(minQ, right)

        forlen(maxQ) > 0 && x >= nums[maxQ[len(maxQ)-1]] {
            maxQ = maxQ[:len(maxQ)-1]
        }
        maxQ = append(maxQ, right)

        // 2. 出：如果窗口不满足要求，左端点离开窗口
        // 只需改下面这行代码，其余逻辑和 2762 题完全一致
        forint64(nums[maxQ[0]]-nums[minQ[0]])*int64(right-left+1) > k {
            left++
            if minQ[0] < left {
                minQ = minQ[1:]
            }
            if maxQ[0] < left {
                maxQ = maxQ[1:]
            }
        }

        // 3. 更新答案
        ans += int64(right - left + 1)
    }
    return
}

func main() {
    nums := []int{1, 3, 2}
    k := 4
    result := countSubarrays(nums, int64(k))
    fmt.Println(result)
}

# -*-coding:utf-8-*-

from collections import deque

def countSubarrays(nums, k):
    ans = 0
    minQ = deque()
    maxQ = deque()
    left = 0
    
    for right, x in enumerate(nums):
        # 1. 入：元素进入窗口
        while minQ and x <= nums[minQ[-1]]:
            minQ.pop()
        minQ.append(right)
        
        while maxQ and x >= nums[maxQ[-1]]:
            maxQ.pop()
        maxQ.append(right)
        
        # 2. 出：如果窗口不满足要求，左端点离开窗口
        while maxQ and minQ and (nums[maxQ[0]] - nums[minQ[0]]) * (right - left + 1) > k:
            left += 1
            if minQ and minQ[0] < left:
                minQ.popleft()
            if maxQ and maxQ[0] < left:
                maxQ.popleft()
        
        # 3. 更新答案
        ans += right - left + 1
    
    return ans

def main():
    nums = [1, 3, 2]
    k = 4
    result = countSubarrays(nums, k)
    print(result)

if __name__ == "__main__":
    main()

#include <iostream>
#include <vector>
#include <deque>

using namespace std;

long long countSubarrays(vector<int>& nums, long long k) {
    long long ans = 0;
    deque<int> minQ, maxQ;
    int left = 0;

    for (int right = 0; right < nums.size(); right++) {
        int x = nums[right];

        // 1. 入：元素进入窗口
        while (!minQ.empty() && x <= nums[minQ.back()]) {
            minQ.pop_back();
        }
        minQ.push_back(right);

        while (!maxQ.empty() && x >= nums[maxQ.back()]) {
            maxQ.pop_back();
        }
        maxQ.push_back(right);

        // 2. 出：如果窗口不满足要求，左端点离开窗口
        while (!minQ.empty() && !maxQ.empty() &&
               (long long)(nums[maxQ.front()] - nums[minQ.front()]) * (right - left + 1) > k) {
            left++;
            if (!minQ.empty() && minQ.front() < left) {
                minQ.pop_front();
            }
            if (!maxQ.empty() && maxQ.front() < left) {
                maxQ.pop_front();
            }
        }

        // 3. 更新答案
        ans += (right - left + 1);
    }

    return ans;
}

int main() {
    vector<int> nums = {1, 3, 2};
    long long k = 4;
    long long result = countSubarrays(nums, k);
    cout << result << endl;
    return0;
}