~~~ /* 時間：2012.03.17 作者：煙大洋仔 題目： 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 題目分析： 最初的時候竟然忘記了數組賦值整了老半天，通過幾個for循環在加上數組的循環判斷即可解決；竟然在這里浪費了半天的時間氣人 */ #include <iostream> using namespace std; int main() { int n,a[100],i=0,sum=0; while (cin>>n&&n!=0) { a[0]=0; sum=0; for(i=1;i<=n;i++) cin>>a[i]; for (i=1;i<=n;i++) { if(a[i]>a[i-1]) sum=sum+(a[i]-a[i-1])*6+5; else sum=sum+(a[i-1]-a[i])*4+5; } cout<<sum<<endl; } return 0; } ~~~