Elevator (20)

1 题目

Elevator (20)
时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB
题目描述
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.

输入描述:
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.

输出描述:
For each test case, print the total time on a single line.

输入例子:
3 2 3 1

输出例子:
41

2 解析

2.1 题意

求乘客乘坐电梯的总共时间花费，单次乘坐的时间分为电梯上升或下降时间，乘客下电梯时间。

2.2 思路

一开始的电梯位于第0层，每上升一层，花费6秒，下降一层楼，花费4秒，在每一层楼下每一名乘客时，需要花费5秒。
注意：如果连续两名乘客到的楼层相同时，后一名的乘客只需要计算下电梯的时间（即5s）。

3 参考代码

#include <cstdio>

int main(int argc, char const *argv[]){
    int n;
    scanf("%d", &n);
    int num, floor = 0, ans = 0;
    for (int i = 0; i < n; ++i) {
        scanf("%d", &num);
        if(num > floor){
            ans += 6*(num - floor) + 5;
        }else if(num < floor){
            ans += 4*(floor - num) + 5;
        }else{
            ans += 5;
        }
        floor = num;
    }
    printf("%d", ans);
    return 0;
}