杭電acm1008
Elevator
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 23 2 3 10
Sample Output
1741
Author ZHENG, Jianqiang
Source ZJCPC2004
Recommend JGShining
實作代碼：
#include <iostream>
using namespace std;
//首先輸入将要輸入的層數的總個數，然後輸入層數，上升一層花費6秒，
//下降一層花費4秒，停在某層花費5秒，0表示結束輸入
int main()
{
int a[200] = {0};
int n,i = 1;
while(cin>>n && 0<=n)
{
int sum =0;
if(n == 0)
break;
for(i=1;i<=n;i++)
{
cin>>a[i];
if(a[i-1] < a[i])
sum += (a[i]-a[i-1])*6+5;
else if(a[i-1] > a[i])
sum += (a[i-1]-a[i])*4+5;
else
sum = sum +5;
}
cout<<sum<<endl;
}
return 0;
}