10055 - Hashmat the Brave Warrior

10055 - Hashmat the Brave Warrior

Time limit: 3.000 seconds

Problem A
Hashmat the brave warrior
Input: standard input
Output: standard output

Hashmat is a brave warrior who with his group of young soldiers moves from one place to another to fight against his opponents. Before fighting he just calculates one thing, the difference between his soldier number and the opponent's soldier number. From
this difference he decides whether to fight or not. Hashmat's soldier number is never greater than his opponent.

Input

The input contains two integer numbers in every line. These two numbers in each line denotes the number of soldiers in Hashmat's army and his opponent's army or vice versa. The input numbers are not greater than 2^32. Input is terminated by End of File.

Output

For each line of input, print the difference of number of soldiers between Hashmat's army and his opponent's army. Each output should be in seperate line.

Sample Input:

10 12

10 14

100 200

[b]Sample Output:[/b]

2

4

100

___________________________________________________________________________________

Shahriar Manzoor

16-12-2000

AC代码：

#include "stdio.h"
int main()
{
long long a ,b ;
while ( EOF != scanf( "%lld%lld" , &a , &b ) )
{
printf( "%lld\n" , a > b ? a - b : b - a ) ;
}
return 0 ;
}

本人刚刚开始进行竞赛训练，只能做些简单的题来找找感觉，以后会慢慢的规范起来。。。

做题反省：

最初没有考虑到int数据范围，之后才用double才A了。

然后，突然觉得数据范围是一个很关键的东西要好好看看才行，不！不是好好看看，而已要牢牢记住！！！

测试 double 如下：

11975345

10055

Hashmat the Brave Warrior

Accepted

ANSI C

0.149

测试 long long 如下：

11975390

10055

Hashmat the Brave Warrior

Accepted

ANSI C

0.066