Hashmat the Brave Warrior

最新推荐文章于 2016-08-15 21:55:26 发布

原创最新推荐文章于 2016-08-15 21:55:26 发布 · 766 阅读

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本文介绍了一个简单的程序挑战，勇士Hashmat会根据自身军队与对手军队的人数差异来决定是否进行战斗。输入包含两组人数，分别代表Hashmat军队与对手军队的人数，输出则是两者之间的差异。

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

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解题报告

该死的编译器，还不能用__int64，题目明明要求a恒小于b，还得判断。

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