杭电OJ Elevator

Elevator

来源： http://acm.hdu.edu.cn/showproblem.php?pid=1008

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35606    Accepted Submission(s): 19443

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
   
   
   
   

 

Author

ZHENG, Jianqiang

 

Source

ZJCPC2004

 

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题解：用数组b[100]存放所要停顿的楼层，n[1000]存放时间。先计算每次电梯停顿的时间，再计算在上下楼时所花费的时间。....

最后输出总时间便是。

#include<iostream>
using namespace std;
int n[1000]={};
int b[100];
int main()
{
	int N,i,j=0;
	cin>>N;
	while(N!=0){
		for(i=0;i<N;i++){
		cin>>b[i];	
		}
		n[j]+=N*5+b[0]*6;
		for(i=0;i<N-1;i++)
		{
			if(b[i]<b[i+1])
			n[j]+=(b[i+1]-b[i])*6;
		   else
		    n[j]+=(b[i]-b[i+1])*4;
		   }
		   j++;  cin>>N;
	}
	for(i=0;n[i]!=0;i++)
	cout<<n[i]<<endl;
	return 0;
}