HDU 1037 Keep on Truckin' (水题）

Keep on Truckin'

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15128    Accepted Submission(s): 10416

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

 

Source

South Central USA 2003

题目大意：有一辆卡车高为168，现有三个桥洞，输入桥洞高度，求是否能通过；三个桥洞都高于168则输出NO CRASH，否则输出CRASH “第一个低于168的高度”。

可以说是非常水的题目了。。。只要能看懂题目就能写出来的那种水题；

不过discuss里面有一个神解，我附在下面了，只能说这题的测试数据可能有问题。

#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;int main(){    int a,b,c;    while(~scanf("%d%d%d",&a,&b,&c)){        if(a>168&&b>168&&c>168){            printf("NO CRASH\n");            continue;        }        if(a<=168)            printf("CRASH %d\n",a);        else if(b<=168)            printf("CRASH %d\n",b);        else if(c<=168)            printf("CRASH %d\n",c);    }    return 0;}

#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;int main(){    printf("CRASH 167\n");    return 0;}