HDU1008——Elevator
2016-08-06 14:57
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Elevator
[align=center]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63914 Accepted Submission(s): 35177
[/align]
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
Print the total time on a single line for each test case.
[align=left]Sample Input[/align]
1 2
3 2 3 1
0
[align=left]Sample Output[/align]
17
41
解:从0层开始,每上升一层用时6秒,下降一层用时4秒,每个楼层停止5秒,则计算上升与下降所用时间,再加上到达指定楼层之后停止时间。
#include<stdio.h> #include<string.h> int a[105]; int main() { int n; while(~scanf("%d",&n) && n){ int t=0; memset(a,0,n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); } t+=a[0]*6; if(n>1){ for(int i=1;i<n;i++){ if(a[i]>a[i-1]){ t+=(a[i]-a[i-1])*6; }else{ t+=(a[i-1]-a[i])*4; } } } t=t+n*5; printf("%d\n",t); } return 0; }
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