HDU1008

Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73498 Accepted Submission(s): 40447

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

Author
ZHENG, Jianqiang

Source
ZJCPC2004

Recommend
JGShining | We have carefully selected several similar problems for you: 1002 1009 1021 1003 1108

---

一道简单的数学计算题，理清计算规则就行，采取边输入边处理可以减少空间开销，但时间开销增大的不明显，没测试。for循环code长度367B没有while的498大，目前不清楚

#include<stdio.h>int main(){    int n, temp, time, floor;    while (scanf("%d", &n)!=EOF,n)    {        temp = time = 0 ;        for(int i=0;i<n;i++)        {            scanf("%d", &floor);            if (floor > temp)            {                time += (floor - temp) * 6 + 5;            }            else            {                time += (temp - floor) * 4 + 5;            }            temp = floor;        }        printf("%d\n",time);    }    return 0;}

#include<stdio.h>int main(){    int n, temp, time, floor;    while (scanf("%d", &n)!=EOF,n)    {        temp = time = 0 ;        while(n--)        {            scanf("%d", &floor);            if (floor > temp)            {                time += (floor - temp) * 6 + 5;            }            else            {                time += (temp - floor) * 4 + 5;            }            temp = floor;        }        printf("%d\n",time);    }    return 0;}