Keep on Truckin'-hdu-1037

Keep on Truckin'

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6805    Accepted Submission(s): 4694

Problem Description

Boudreaux and Thibodeaux are on the road again . . .

"Boudreaux, we have to get this shipment of mudbugs to Baton Rouge by tonight!"

"Don't worry, Thibodeaux, I already checked ahead. There are three underpasses and our 18-wheeler will fit through all of them, so just keep that motor running!"

"We're not going to make it, I say!"

So, which is it: will there be a very messy accident on Interstate 10, or is Thibodeaux just letting the sound of his own wheels drive him crazy?

 

 

Input

Input to this problem will consist of a single data set. The data set will be formatted according to the following description.

The data set will consist of a single line containing 3 numbers, separated by single spaces. Each number represents the height of a single underpass in inches. Each number will be between 0 and 300 inclusive.

 

 

Output

There will be exactly one line of output. This line will be:

   NO CRASH

if the height of the 18-wheeler is less than the height of each of the underpasses, or:

   CRASH X

otherwise, where X is the height of the first underpass in the data set that the 18-wheeler is unable to go under (which means its height is less than or equal to the height of the 18-wheeler).
The height of the 18-wheeler is 168 inches.

 

 

Sample Input

180 160 170

 

 

Sample Output

CRASH 160

中文翻译： 

问题说明
德留克斯和THIBODEAUX的的道路上再次。 。 。

“包德留克斯，我们必须得到这个mudbugs装运，巴吞鲁日，在今晚！”

“不要担心，THIBODEAUX，我已经提前检查​​有三个地下通道，我们18轮将通过所有这些，所以只是保持，电机运行！”

“我们不会做它，我说！”

所以，这是它会有一个非常混乱的10号州际公路上的事故，或者是THIBODEAUX只是让自己的车轮声开车送他疯了吗？

 

输入
输入到这个问题，将包括一个单一的数据集。根据下面的描述中，该数据集将被格式化。

该数据集将包括一行含有3个号码，用一个空格隔开。每个数字代表一个单一的地下道英寸的高度。每个号码可获介于0和300（含）。

 

产量
将是完全有一行输出。这条线将是：

   没有崩溃

如果在18轮的高度小于在地下通道中的每一个的高度，或

   CRASH X

否则，其中X是第一个地下通道中的数据集，18轮无法去下（这意味着它的高度是小于或等于18轮的高度）的高度。
18轮的高度为168英寸。
#include<stdio.h>

int main()

{

    int n,i,a[3];

    while(scanf("%d%d%d",&a[0],&a[1],&a[2])!=EOF)

    {

      for(i=0;i<3;i++)

      {

       if(a[i]<=168)

       {printf("CRASH %d\n",a[i]);break;}

      }

      if(i==3)

      printf("NO CRASH\n");

    }

    return 0;

}//不要想的太难，其实就是找小于等于168的数，找到一个即可。。如果都大于168 则不崩溃。