HDOJ1008 Elevator
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Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1593 Accepted Submission(s): 807
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.
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 23 2 3 10
Sample Output
1741
Source
ZHENG, Jianqiang
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PS:本来做这个题是为了给自己打打气,增加一点自信,因为看到1006 Tick and Tick没有一点思路,可是.......竟然WA了二次
/**//*
Name:
Copyright:
Author:
Date: 26-07-07 17:27
Description: 一定注意初始化的问题,一般的题目都要求输入N个input
还要注意的是考虑相邻二个命令都在一个楼层
*/
#include<iostream>
using namespace std;
int main()
...{
int N;
while(cin>>N)
...{
int time=0,from=0,to=0;
if(N==0) break;
for(int i=0;i<N;i++)
...{
cin>>to;
if(to>from)
time+=(to-from)*6+5;
else
time+=(from-to)*4+5;
from=to;
}
cout<<time<<endl;
}
}
Name:
Copyright:
Author:
Date: 26-07-07 17:27
Description: 一定注意初始化的问题,一般的题目都要求输入N个input
还要注意的是考虑相邻二个命令都在一个楼层
*/
#include<iostream>
using namespace std;
int main()
...{
int N;
while(cin>>N)
...{
int time=0,from=0,to=0;
if(N==0) break;
for(int i=0;i<N;i++)
...{
cin>>to;
if(to>from)
time+=(to-from)*6+5;
else
time+=(from-to)*4+5;
from=to;
}
cout<<time<<endl;
}
}
Download the code
http://dl2.csdn.net/down4/20070726/26174626265.cpp
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