小白學JAVA-----P1008

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 23 2 3 10

 

Sample Output

1741

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import java.util.Scanner;

public class Main {


public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int N=sc.nextInt();//要去几个楼层
if(N==0){
break;
}
int a[]=new int[N];//定义一个数组，把要去的楼层装在一个数组里
for(int i=0;i<N;i++){
a[i]=sc.nextInt();//输入每一个楼层
}

int t=0;
t=N*5;//在每一层都要停5秒
for(int i=0;i<N;i++){
if(i==0){
t=t+a[i]*6;
}else{
if(a[i]>a[i-1]){
t=t+(a[i]-a[i-1])*6;
}

if(a[i]<a[i-1]){
t=t+(a[i-1]-a[i])*4;
}
}

}
System.out.println(t);
}
}

}

英语不好的这道题难度很大 因为读不懂题目的意思，这道题意思是：刚才在第0层处于停滞状态，上一层楼需要6秒，下一层楼需要4秒，到指定楼层停5秒。例如

1 2  就是从0层开始上了两层就是2*6=12秒加上在0层停的5秒就是17秒3 2 3 1  这里就是从0到2用了2*6=12秒加上从2到3的6秒加上从3到1的8秒加上每到一层停的5秒就是12+6+8+15=41秒