本文分享的必刷题目是从蓝桥云课、洛谷、AcWing等知名刷题平台精心挑选而来并结合各平台提供的算法标签和难度等级进行了系统分类。题目涵盖了从基础到进阶的多种算法和数据结构旨在为不同阶段的编程学习者提供一条清晰、平稳的学习提升路径。欢迎大家订阅我的专栏算法题解C与Python实现附上汇总贴算法竞赛备考冲刺必刷题C | 汇总【题目来源】AtCoderE - Temperature Fluctuation Range【题目描述】Takahashi is analyzing weather data. There is a record of temperature observations taken overN NNconsecutive days in a certain region, where the temperature on dayi iiwasH i H_iHi​degrees.高桥正在分析天气数据。某个地区记录了连续N NN天的气温观测值其中第i ii天的气温为H i H_iHi​度。Takahashi wants to select a consecutive period ofK KKdays from this observation data and investigate the temperature variation range during that period.高桥希望从这些观测数据中选取连续K KK天的时间段并研究该时间段内的气温变化范围。Here, the “temperature variation range” for a consecutive period ofK KKdays is defined as the difference between the maximum temperature and the minimum temperature during that period.此处连续K KK天时间段的气温变化范围定义为该时间段内最高气温与最低气温的差值。Takahashi wants to find a consecutive period ofK KKdays that maximizes the temperature variation range. Find the maximum value of the temperature variation range.高桥希望找到一个连续K KK天的时间段使得气温变化范围最大。求气温变化范围的最大值。【输入】N NNK KKH 1 H 2 … H N H_1\ H_2\ \dots\ H_NH1​H2​…HN​The first line containsN NN, representing the number of observation days, andK KK, representing the number of consecutive days to select, separated by a space.The second line containsH 1 , H 2 , … , H N H_1,H_2,…,H_NH1​,H2​,…,HN​, representing the temperature on each day, separated by spaces.【输出】Output the maximum value of the temperature variation range in one line.【输入样例】5 3 2 5 1 8 4【输出样例】7【解题思路】【算法标签】#单调队列#【代码详解】#includebits/stdc.husingnamespacestd;#defineintlonglongconstintN200005;intn,k;// n: 数组长度k: 滑动窗口大小inta[N];// 原始数组intans-2e18;// 存储最大差值初始化为极小值structNode{intv,idx;// v: 数值idx: 索引};dequeNodeminq,maxq;// 双端队列minq维护窗口最小值maxq维护窗口最大值signedmain(){cinnk;// 读入数组长度和窗口大小for(inti1;in;i){cina[i];// 读入数组元素}for(inti1;in;i)// 遍历数组{// 维护minq队列移除窗口外的元素while(!minq.empty()minq.front().idxi-k){minq.pop_front();}// 维护maxq队列移除窗口外的元素while(!maxq.empty()maxq.front().idxi-k){maxq.pop_front();}// 维护minq队列保持队列单调递增while(!minq.empty()minq.back().va[i]){minq.pop_back();}// 维护maxq队列保持队列单调递减while(!maxq.empty()maxq.back().va[i]){maxq.pop_back();}// 将当前元素加入两个队列minq.push_back({a[i],i});maxq.push_back({a[i],i});// 当窗口大小达到k时计算当前窗口的最大差值if(ik){ansmax(ans,maxq.front().v-minq.front().v);}}coutansendl;// 输出结果return0;}【运行结果】5 3 2 5 1 8 4 7