Elevator(电梯)

时间：2015-05-06 15:10:09 阅读：103 评论：0 收藏：0 [点我收藏+]

标签：

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50649 Accepted Submission(s): 27934

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

#include<stdio.h>

main()
{
int a,n,b,c,d,f,j,i,s[109];
while(scanf("%d",&n)&&n!=0)
{b=n;
//c=0;
for(i=0;i<b;i++)
scanf("%d",&s[i]);
c=s[0]*6+5;
for(i=0;i<b-1;i++)
{
if(s[i+1]>s[i])
c=c+(s[i+1]-s[i])*6+5;
else
c=c+(s[i]-s[i+1])*4+5;
}
printf("%d\n",c);
}
return 0;
}

Elevator(电梯)

标签：

原文地址：http://blog.csdn.net/l15738519366/article/details/45535309

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行