PAT甲级刷题记录——1008 Elevator (20分)

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 Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

41

思路

这题很简单，就是模拟一下就行了。

• 如果目标层数比当前层数高，那么需要上楼，于是时间就要加上6×（目标层数-当前层数），并停留5s，更新当前层数；
• 如果目标层数比当前层数低，那么需要下楼，于是时间就要加上4×（当前层数-目标层数），并停留5s，更新当前层数；
• 如果相等（这种情况是比较坑的，题目都没说清楚），依然要停留5s。

代码

#include
#include
#include
#include
#include
using namespace std;
int main()
{
     
    int N;
    scanf("%d", &N);
    int nowFloor = 0;
    int sum = 0;
    for(int i=0;i<N;i++){
     
        int tmpFloor;
        scanf("%d", &tmpFloor);
        if(tmpFloor>nowFloor){
     
            sum += 6*(tmpFloor-nowFloor);
            sum += 5;
        }
        else if(tmpFloor<nowFloor){
     
            sum += 4*(nowFloor-tmpFloor);
            sum += 5;
        }
        else{
     
            sum += 5;
        }
        nowFloor = tmpFloor;
    }
    printf("%d", sum);
    return 0;
}