HDOJ 1008 Elevator 解题报告
下午回来写HDOJ 1008,
一看题目觉得真的很简单,,
写着写着,,
写完了突然发现AC不了,,
调试之后发现还是没问题,
但就是AC不了。
无奈之下看了别人写的代码,
原来是自己题目理解错误了,
题目的意思是到了该楼层再停,,
我误以为是每层都停下来,,
所以写错了。。。。
这题确实是一道水题。。
http://acm.hdu.edu.cn/showproblem.php?pid=1008
Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32135 Accepted Submission(s): 17432
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.
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
Recommend
JGShining
没什么好说的,真的水题,
按照题目意思来做,
就顺利写出来了。
#include "stdio.h" int main() { int i,n; int t[100],sum; while (scanf("%d",&n)!=EOF &&n !=0) { t[0]=0; for(i=1, sum=0 ;i<=n;i++) { scanf("%d",&t[i]); if(t[i-1]<t[i]) sum+=(t[i]-t[i-1])*6+5; //上楼 else sum+=(t[i-1]-t[i])*4+5; //下楼 } printf("%d\n",sum); } return 0; }