1006 of strategy
2016-03-25 23:23
225 查看
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
Print the total time on a single line for each test case.
Sample Input
1 2
3 2 3 1
0
Sample Output
17
41
题目要求:提供电梯要经过的楼层,算出运行的时间。
思路:直接进行上6下4停5的过程进行计算。
细节:没有什么坑,又是一道签到题。
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
Print the total time on a single line for each test case.
Sample Input
1 2
3 2 3 1
0
Sample Output
17
41
题目要求:提供电梯要经过的楼层,算出运行的时间。
思路:直接进行上6下4停5的过程进行计算。
细节:没有什么坑,又是一道签到题。
#include<iostream> #include<cstdio> using namespace std; int main() { int i,j,n,sum=0; int a[101]={0}; while(scanf("%d",&n)) { if(n==0)break; j=n;sum=5*n; while(n--) scanf("%d",&a[n+1]); for(i=j-1;i>0;i--) if(a[i]>a[i+1])sum+=(a[i]-a[i+1])*6; else sum+=(a[i+1]-a[i])*4; sum+=a[j]*6; printf("%d\n",sum); } }
相关文章推荐
- 闭关日记 Day13
- 【游戏服务器开发】SDK接入——以nibia为例的SDK接入总结
- 弦月下的SQL<3> 表的创建
- 字符串小知识
- 多线程:图片下载案例
- 80端口被占用(端口检查)解决--转
- 推荐一个免费的论文查重网站
- Binary Tree Inorder Traversal
- 多线程:GCD 基本使用
- java 递归
- 弦月下的SQL<2> 数据库管理
- CentOS-6.5下编译安装LNMP环境【nginx1.8.1、mysql5.5.43、php 5.6】
- key word
- 游标使用效率对比
- 多线程:block 演练
- Spring事务管理(详解+实例)
- 职场“潜”规则:心法和技法
- SQL语言简介
- zk-002 zookeeper的基本概念
- 多线程:UIView 动画案例