本题要求编写程序，计算2个有理数的和、差、积、商。

输入格式：

输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为0。

输出格式：

分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”，其中k是整数部分，a/b是最简分数部分；若为负数，则须加括号；若除法分母为0，则输出“Inf”。题目保证正确的输出中没有超过整型范围的整数。

输入样例1：
2/3 -4/2
输出样例1：
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
输入样例2：
5/3 0/6
输出样例2：
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

解题：简单的分数运算，可能会出错的地方是如果没有用long long可能在第三四的数据点WA，此外此题要用到辗转相除法，否则会在第四个检测点超时。

代码：

include

include

include<string.h>

using namespace std;

long long gcd(long a, long b)
{
if (b == 0) return a;
return gcd(b, a%b);
}

void deal(long long a, long long b)
{
if (b == 0) cout << "Inf";
else if (a%b == 0)
{
if (a / b >= 0)
{
cout << a / b;
}
else cout << "(" << a / b << ")";
}
else
{
long long fu = a / b;
a -= fu*b;
long long m = a, n = b;
if (m < 0) m = 0 - m;
if (n < 0) n = 0 - n;
long long s = gcd(m, n);
a /= s;
b /= s;
if (fu < 0)
{
if (a < 0) a = 0 - a;
cout << "(" << fu << " " << a << "/" << b << ")";
}
else if (fu != 0)
{
if (a < 0) a = 0 - a;
cout << fu << " " << a << "/" << b;
}
else
{
if (b < 0)
{
b = 0 - b;
a = 0 - a;
}
if (a < 0)
{
cout << "(" << a << "/" << b << ")";
}
else
cout << a << "/" << b;
}
}
}

int main()
{
char q = 0;
long long a, b, c, d;
cin >> a >> q >> b >> c >> q >> d;
long long jin1 = 0, jin2 = 0;
char s[4] = { '+','-','_','/' };
for (long long i = 0; i< 4; i++)
{
deal(a, b);
cout << " " << s[i] << " ";
deal(c, d);
cout << " = ";
switch (i)
{
case 0:
deal(a_d + c_b, b_d);
cout << endl;
break;
case 1:
deal(a_d - c_b, b_d);
cout << endl;
break;
case 2:
deal(a_c, b_d);
cout << endl;
break;
case 3:
if (c < 0)
{
a = 0 - a;
c = 0 - c;
}
deal(a_d, b*c);
break;
}
}

}