[结题报告]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

 

Sample Output:

2
4
100

 

 

 

代码参考：

本题大意求Hashmat与敌人的数量差，值得注意的是无论Hashmat的人数多余敌人还是少于敌人,输出均为正数.

#include"stdio.h"
main()
{ double a,b; //定义变量； 
while(scanf("%lf%lf",&a,&b)!=EOF) //输入 
{if(a<b) printf("%.0lf\n",b-a); //判断,保证输出为正数 
else printf("%.0lf\n",a-b);
}
}