电梯的c语言数组题目，求解

#include 
#define MAX 30
int main()
{
    int n, pf, f, t=0;

    pf = f = 0;
    scanf("%d", &n);
    while(n--)
    {
        while(1)
        {
            scanf("%d", &f);
            if(f>MAX)
                printf("Too big floor number, input it again.\n");
            else
                break;
        }

        if(f >= pf) //上行
            t += (f-pf)*6+5;
        else        //下行
            t += (pf-f)*4+5;
        pf = f;
    }
    printf("%d",t);
}

int main()
{
	//此处忽略将输入读取到数组中的步骤
	int floors[] = {2,3,1}; //输入楼层数据
	int allTimes = 0; //设定总消耗时间

	//设定电梯前一次所在楼层，初始为0
	int lastFloor = 0;

	//开始跟随电梯遍历每个楼层
	int arrayCount = sizeof(floors)/sizeof(int);//计算数组长度
	for(int i=0;i		//每次移动产生的时间消耗为移动的时间+1次停留消耗的时间
		int s = floors[i]-lastFloor;
		if(s>0){ //上升
			allTimes += s*6 + 5;
		}else {
			allTimes += s*4 + 5;
		}
	}

	printf("最终消耗的总时间为：%d",allTimes);

}

//注意：本处没有考虑 错误和异常输入的检测，也没有进行常量的提取，方法的模块化等工作，只是给出一个完整的实现思路。代码可运行！

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