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1 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 = Infps：
#define _CRT_SECURE_NO_WARNINHS
#define _CRT_SECURE_NO_DEPRECATE#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
#include<numeric>
#include<functional>
#include<vector>
#include<stack>using namespace std;int gcd(long long a, long long b)
{return b == 0 ? a : gcd(b, a % b);
}void print(long long a, long long b)
{long long c = 0; if(a > 0){ if(b == 1){ printf("%lld", a);}else if(a > b){ c = a / b;a -= b * c;printf("%lld %lld/%lld", c, a, b);}else{ printf("%lld/%lld", a, b);}}else if(a == 0){printf("%c", '0');}else{ if(b == 1){ printf("(%lld)", a);}else if(-1 * a > b){ c = a / b;a = (-1 * a) % b;printf("(%lld %lld/%lld)", c, a, b);}else{ printf("(%lld/%lld)", a, b);}}
}void add(long long a1, long long b1, long long a2, long long b2);
void subtract(long long a1, long long b1, long long a2, long long b2);
void multiply(long long a1, long long b1, long long a2, long long b2);
void divide(long long a1, long long b1, long long a2, long long b2);int main()
{long long a1, b1, a2, b2;scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);long long gcd1 = abs(gcd(a1, b1));a1 /= gcd1;b1 /= gcd1;long long gcd2 = abs(gcd(a2, b2));a2 /= gcd2;b2 /= gcd2;add(a1, b1, a2, b2);subtract(a1, b1, a2, b2);multiply(a1, b1, a2, b2);divide(a1, b1, a2, b2);return 0;
}void add(long long a1, long long b1, long long a2, long long b2)
{print(a1, b1);printf(" + ");print(a2, b2);printf(" = ");long long a3 = a1 * b2 + a2 * b1;long long b3 = b1 * b2;long long gcd3 = abs(gcd(a3, b3));a3 /= gcd3;b3 /= gcd3;print(a3, b3);printf("\n");
}void subtract(long long a1, long long b1, long long a2, long long b2)
{print(a1, b1);printf(" - ");print(a2, b2);printf(" = ");long long a3 = a1 * b2 - a2 * b1;long long b3 = b1 * b2;long long gcd3 = abs(gcd(a3, b3));a3 /= gcd3;b3 /= gcd3;print(a3, b3);printf("\n");
}void multiply(long long a1, long long b1, long long a2, long long b2)
{print(a1, b1);printf(" * ");print(a2, b2);printf(" = ");long long a3 = a1 * a2;long long b3 = b1 * b2;long long gcd3 = abs(gcd(a3, b3));a3 /= gcd3;b3 /= gcd3;print(a3, b3);printf("\n");
}void divide(long long a1, long long b1, long long a2, long long b2)
{print(a1, b1);printf(" / ");print(a2, b2);printf(" = ");if(a2 == 0){printf("Inf");}else if(a2 < 0){long long a3 = -1 * a1 * b2;long long b3 = -1 * b1 * a2;long long gcd3 = abs(gcd(a3, b3));a3 /= gcd3;b3 /= gcd3;print(a3, b3);}else{long long a3 = a1 * b2;long long b3 = b1 * a2;long long gcd3 = abs(gcd(a3, b3));a3 /= gcd3;b3 /= gcd3;print(a3, b3);}printf("\n");
}