2016SDAU课程练习一 1006

Problem G

Problem DescriptionThe 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

题意：一个电梯，下一层4秒，上一层6秒，一层停留5秒，电梯从0层开始，给一个楼层n，多长时间能到

思路：这个题理解起来比较简单，如果后一个楼层大于前一个，就上，否则就下，再加上停留时间即可

感想：没发现这题贪在哪里。。。

AC代码：

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