10055-Hashmat the brave warrior

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

解题思路：

该题乍一看是很简单，但依旧有陷阱，题中有“or vice versa”即相反输入也可，所以我们需要取两数相减的绝对值而不是后面的数减前面的数。

另输入数据大小不超过2^32，int的范围是-2^31~2^31-1，所以我们需要用long long类型来存储数据，否则无法AC。

AC代码：

#include<iostream>#include<cstdio>#include<cmath>using namespace std;int main(){    long long a, b, answer;    while( scanf("%lld%lld", &a, &b) != EOF )    {        answer = fabs (a - b);//求出两军队数量之差的绝对值        printf("%lld\n",answer);    }    return 0;}