贪心算法2之1006

1 题目编号：1006  problemG

2 题目内容：

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

3  解题思路形成过程：

此题我认为最重要的是正确理解题意。此题我一共有3种理解，一为第一次升降电梯后其始点变为第一次结束时的楼层，此后以此类推结果证明是错误的。二为同一楼层时不会耗费时间，结果证明有是错误的，三即正确思路了。同一楼层也会消耗5min，

4 感想：认真想，画画过程图，即可模拟出过程。

5 代码：

#include<iostream>
#include<stdlib.h>
//#include<algorithm>
using namespace std;
int main()
{
//int iTime = 0;
int N;
while (cin >> N)
{
int iTime = 0;
int iStart = 0;
int iStopNum[101];
if (N == 0)
break;
for (int i = 0; i < N; i++)
{
cin >> iStopNum[i];
if (iStopNum[i]>iStart)
{
iTime += (iStopNum[i] - iStart) * 6 + 5;
iStart = iStopNum[i];
}
else if (iStopNum[i] < iStart)
{
iTime += (iStart - iStopNum[i]) * 4 + 5;
iStart = iStopNum[i];
}
else
{
iTime += 5;
}
}
cout << iTime << endl;
}
return 0;
}