题目大意:就是说现在有个升降机，上升一层就要花6秒时间，到了目标楼层后会停留5秒时间，下降一层要花4秒时间，你一开始现在在0层，OJ提供你楼层信息，你计算出整个过程所花费的时间。 |
#include<stdio.h>
#include<string.h>
#define Max 105
int main()
{
int floor[Max],N,i;
int temp;
memset(floor,0,sizeof(floor));
while(scanf("%d",&N)&&N)
{
int time=0;
for(i=1;i<=N;i++)
{
scanf("%d",&floor[i]);
temp=floor[i]-floor[i-1];
if(temp>0)
{
time+=temp*6+5;
continue;
}
if(temp<0)
{
time+=-temp*4+5;
continue;
}
if(temp==0)
{
time+=5;
continue;
}
}
printf("%d\n",time);
}
return 0;
}
ElevatorTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46694 Accepted Submission(s): 25735
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 |