GDUFE ACM-1045

最新推荐文章于 2025-08-15 22:56:20 发布

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最新推荐文章于 2025-08-15 22:56:20 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ruo786828164/p/6009175.html

本文针对一项经典的电梯移动问题进行了算法解析。该问题涉及电梯在不同楼层间的移动与停留时间计算，通过对输入楼层序列的处理，实现了高效的时间计算方法。

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题目：http://acm.gdufe.edu.cn/Problem/read/id/1045

Elevator

Time Limit: 2000/1000ms (Java/Others)

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

思路：每输入一个数字都要加5秒，判断输入的数字与前一个数字哪个比较大，从而得知加6秒还是4秒

难度：简单

代码：

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int i,a,c;
 5     int b[100];
 6     while(~scanf("%d",&a))
 7     {
 8         if(a==0) break;
 9         for(i=0;i<a;i++)
10             scanf("%d",&b[i]);
11         c=b[0]*6+5;
12         for(i=1;i<a;i++)
13         {
14             if(b[i]<b[i-1])
15                 c=c+(b[i-1]-b[i])*4;
16             if(b[i]>b[i-1])
17                 c=c+(b[i]-b[i-1])*6;
18                 c=c+5;
19         }
20         printf("%d\n",c);
21     }
22     return 0;
23 }