【模拟】HDU 1008 Elevator【题解】

Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 71305 Accepted Submission(s): 39169

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

Author
ZHENG, Jianqiang

题意

这个题的意思就是你要坐电梯，这个电梯每向上前进一层需要花费6s，下降一层需要花费4s。停靠的时候还需要花费5s（大概是用来开门吧）
例如：
从0到4 需要花费 6*（4-0）+5 也就是29s。
但是比较坑爹的是，即使在同一层，比如从4到4 还需要花5s。(⊙o⊙)

AC代码：

#include<cstdio>
#include<iostream>

long long int sum;

void add(int a,int b){
    if(a<b)
        sum+= (b-a)*6+5;
    else
        sum+= (a-b)*4+5;
}

int main(){
    int tmp1,tmp2;
    int n;
    while(~scanf("%d",&n)&&n){
        sum=tmp2=0;
        for(int i=0;i<n;++i){
            tmp1=tmp2; /*这道题没必要用数组存储，两个变量就可以完成所有的录入。*/
            scanf("%d",&tmp2);
            add(tmp1,tmp2);
        }
        printf("%lld\n",sum);
    }
}