[ACM]Elevator

发布时间:2016-12-6 10:48:55 编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"[ACM]Elevator",主要涉及到[ACM]Elevator方面的内容,对于[ACM]Elevator感兴趣的同学可以参考一下。

Problem Description The 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 There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed. Output Sample Input 1 2 3 2 3 1 0 Sample Output 17 41 Author ZHENG, Jianqiang Source ZJCPC2004 解题思路: 开一个数组用来保存输入的楼层数,变量time为总共的时间,初始化为0,从数组的第个数开始,依次比较当前数与前一个数的大小,分情况time+=(6或4)*(大数-小数),最后time再加上数组中的一个数(楼层数)所用的时间,再加上所有楼层停留的时间。 代码: #include <iostream> using namespace std; int fl[102]; int main() { int n; while(cin>>n&&n) { int i; for(i=0;i<n;i++) cin>>fl[i]; int time=0; for(i=1;i<n;i++) { if(fl[i]>fl[i-1])//比较 time=time+6*(fl[i]-fl[i-1]); if(fl[i]<fl[i-1]) time=time+4*(fl[i-1]-fl[i]); } cout<<time+6*fl[0]+5*n<<endl; } return 0; } 运行截图:

上一篇:第一个IOS app- 密码管理大师
下一篇:boost库asio详解3——io_service作为work pool

相关文章

关键词: [ACM]Elevator

相关评论