PAT (Advanced Level) Practise:1008. Elevator
2015-09-01 11:01
357 查看
【题目链接】
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 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:
Sample Output:
提交代码:
运行结果:
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 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 1
Sample Output:
41
提交代码:
#include <stdio.h> int main(void) { int N; int i, pre, cur; int delta; int time; pre = 0; time = 0; scanf("%d", &N); for(i = 0; i < N; i++) { scanf("%d", &cur); if(cur - pre >= 0) { delta = cur - pre; time += (delta * 6); } else { delta = pre - cur; time += (delta * 4); } pre = cur; } printf("%d", time + N * 5); return 0; }
运行结果:
相关文章推荐
- 325 zb的生日【dfs】
- vim使用说明
- 栈的基本操作
- HDU 2594(Simpsons’ Hidden Talents)字符串匹配-KMP
- HDU 2081 手机短号
- CVPR 2015 open access
- Ambari Metrics介绍
- 在Runbook中添加Checkpoint-workflow
- nginx 不带www到www域名的重定向
- 关于多线程三种语句的实现
- MarkdownPad2注册码
- Android XML动画资源文件详细讲解(上)
- Jenkins+Gradle实现android开发持续集成问题汇总
- nyoj 844 A+B Problem(V)【long long】
- 基于Cef的简易浏览器开发(CefSharp)
- C语言结构体(struct)常见使用方法
- nslog 判断打印的 那一行
- TCP连接
- #3 working with data stored in files && securing your application
- SynchronousQueue的简单应用2