Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 64178 Accepted Submission(s): 35350
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
水的一笔的题 看懂就好
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #define N 100 6 using namespace std; 7 int a[N]; 8 int main() 9 { 10 int n; 11 while(scanf("%d",&n)!=EOF) 12 { 13 if(n==0) 14 break; 15 int sum=n*5; 16 memset(a,0,sizeof(0)); 17 int i; 18 for(i=1;i<=n;i++) 19 { 20 scanf("%d",&a[i]); 21 } 22 int j; 23 a[0]=0; 24 for(i=0;i<n;i++) 25 { 26 j=i+1; 27 if(a[j]>a[i]) 28 sum=sum+abs(a[j]-a[i])*6; 29 if(a[j]<a[i]) 30 sum=sum+abs(a[j]-a[i])*4; 31 } 32 printf("%d ",sum); 33 } 34 }