【PAT甲级真题】- Elevator(20)

题目来源Elevator题目描述点击链接自行查看注意点停在同一层时多等5秒DescriptionThe highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.Input Specification:Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.Output Specification:For each test case, print the total time on a single line.Sample Input:3 2 3 1Sample Output:41题目大意给出N个请求楼层电梯从第 0 层开始按照请求的顺序到达指定楼层每上一层花费 6 秒下一层花费 4 秒为每个请求停留 5 秒钟计算满足所有请求的总时间思路简介一道很简单的模拟题按照题目的意思模拟即可用一个 last 暂存上一步的状态根据当前输入判断上升还是下降移动了几层然后乘上时间输出时加上请求数目乘上 5一开始怀疑是不是如果一直重复同一个请求就都只用 5 秒看来实际情况是每个请求不管是不是同一层都会花费 5 秒钟等待遇到的问题无一遍过代码/** * https://pintia.cn/problem-sets/994805342720868352/exam/problems/type/7?problemSetProblemId994805511923286016 * 模拟 */#includebits/stdc.husingnamespacestd;voidsolve(){intn;cinn;intlast0,res5*n;for(inti0;in;i){intnow;cinnow;intmovenow-last;if(move0)resmove*6;elseres-move*4;lastnow;}coutres\n;}intmain(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//fstream in(in.txt,ios::in);cin.rdbuf(in.rdbuf());intT1;//cinT;while(T--){solve();}return0;}