PAT A1008(无脑题)

最新推荐文章于 2021-08-18 12:42:21 发布

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本文介绍了一个关于电梯在不同楼层间移动的时间计算问题。电梯从0楼开始，根据输入的楼层列表进行移动，向上移动每层花费6秒，向下移动每层花费4秒，并在每个楼层停留5秒。文章提供了完整的C++代码实现。

1008. Elevator (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

#include<cstdio>

int main(){
    int num;
    scanf("%d",&num);
    int front = 0;
    int sum = 0;
    for(int i=0;i<num;i++){
        int now;
        scanf("%d",&now);
        if(now>front){
            sum+=(now-front)*6+5;
        }else{
            sum+=(front-now)*4+5;
        }
        front = now;
    }
    printf("%d",sum);
    return 0;
}