我哪错了、老是AC不了。谢谢各位大仙、

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
 

#include <stdio.h>
int main()
{
    char a[1000];
    int i,c,n;
    scanf("%d",&n);
    while(n!=0)
    {
    n++;
    a[0]='0';

    //输入
    for(i=1;i<n;i++)
    {
    scanf("%c",&a[i]);
    if(a[i]==' ') 
    {
        i--;
    }
    if(a[i]=='\n')
    {
        a[i]='\0';
        break;
    }
    }
    //计算
    
        c=0;
        for(i=1;i<n;i++)
        {
            if(a[i]-a[i-1]>0)
            {
                c+=(a[i]-a[i-1])*6+5;

            }
            else if(a[i]-a[i-1]<0)
            {
                c+=(a[i-1]-a[i])*4+5;

            }
            else
            {
                c+=5;
            }
        }
        
    printf("%d\n",c);
    scanf("%d",&n);
    }

    

    return 0;
}

#include <stdio.h>

int main() {
    int n, a[110], sum, i;
    while(scanf("%d", &n) && n != 0) {
        sum = 0;
        for(i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(i = 1, a[0] = 0; i <= n; i++) {
            if(a[i] > a[i - 1]) sum += (a[i] - a[i - 1]) * 6 + 5;
            else sum += (a[i - 1] - a[i]) * 4 + 5;
        }
        printf("%d\n", sum);
    }
    return 0;
}

楼主用字符窜类型，表面上节省空间，但是在输入上参生很大的困难。所以楼上版主的，用整型有利于与数据输入，不过感觉上面版主的程序有一点小问题，就是到大目标层的时候，停的时间应该是不算的，但楼上版主算了。看后一点点想法，小白水准，若说错，望见谅

呵呵,我想你想多了，对于int型来说，应该已经算大了....而且题目是输入的层次的时间是暂停的...还有，我想你没仔细看题吧...最高不超过100层..

[ 本帖最后由 Susake 于 2013-4-6 22:13 编辑 ]