[UvaOJ - Getting Started] 10055 - Hashmat the Brave Warrior

Problem A

Hashmat the brave warrior

Input: standard input

Output: standard output

 

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

Input

The input contains two integer numbers inevery line. These two numbers in each line denotes the number of soldiers inHashmat's army and his opponent's army or vice versa. The input numbers are notgreater 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 hisopponent's army. Each output should be in seperate line.

 Sample Input:

10 12
10 14
100 200

 Sample Output:

2
4
100

___________________________________________________________________________________
Shahriar Manzoor

16-12-2000

程序要点：

The input numbers are notgreater than 2^32. ： 在声明变量的时候要考虑类型 2 ^ 32 需要声明为long long

Input is terminated by End of File. ： 需要判断scanf的返回值类型为EOF

代码：

C语言: Codee#26451

01 #include <stdio.h>
02
03 int main()
04 {
05     long long hashmat = 0 ;
06     long long enemy = 0;
07
08     while ( scanf( "%lld %lld" , & hashmat , & enemy) != EOF);
09     {
10         if ( hashmat >= enemy)
11         {
12             printf( "%lld" , ( hashmat - enemy));
13         }
14         else
15         {
16             printf( "%lld" , ( enemy - hashmat));
17         }
18     }
19
20     return 0;
21 }