题意：电梯上升一楼要6秒，下降一楼要5秒，停在当前层要5秒，开始电梯在0层，然后给出一个升降表，求完成这个表所需要的时间。 之前没注意数据格式，把我都弄晕了。。。（提醒自己注意格式）

Elevator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 52712 Accepted Submission(s): 29123

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

ac代码

#include <cstdio>
#include <cstring>
#define maxn 100
int f[maxn+1];
int main(){
int n;
while(scanf("%d",&n)!=EOF&&n!=0){
memset(f,0,sizeof(f));
int t=0;
for(int i=1;i<=n;i++){
scanf("%d",&f[i]);
int temp=f[i]-f[i-1];
if(temp>0)
t+=temp6+5;
else if(temp<0)
t+=-temp4+5;
else
t+=5;
}
printf("%d\n",t);
}
return 0;
}