# 题目

2/3 -4/2

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

5/3 0/6

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

# 分析

1.输出单个分数的方法；
2.求最大公约数的方法；（这个一定要写的，没用到肯定是有问题，最小公倍数可以偷懒不写）
3.加法；
4.减法；
5.乘法；
6.除法。

# 源代码

//C/C++实现
#include <iostream>
#include <cmath>

using namespace std;

//辗转相除法
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; //带分数前面的整数部分，默认是0
if(a > 0){ //正数
if(b == 1){ //形如3/1
printf("%lld", a);
}
else if(a > b){ //形如5/3
c = a / b;
a -= b * c;
printf("%lld %lld/%lld", c, a, b);
}
else{ //真分数 形如3/5
printf("%lld/%lld", a, b);
}
}
else if(a == 0){ //形如0/3
printf("%c", '0');
}
else{ //负数
if(b == 1){ //形如-3/1
printf("(%lld)", a);
}
else if(-1 * a > b){ //形如-5/3
c = a / b;
a = (-1 * a) % b;
printf("(%lld %lld/%lld)", c, a, b);
}
else{ //真分数
printf("(%lld/%lld)", a, b);
}
}
}

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;
//化简到最简形式，非带分数形式
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;
//化简到最简形式，非带分数形式
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;
//化简到最简形式，非带分数形式
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;
//化简到最简形式，非带分数形式
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;
//化简到最简形式，非带分数形式
long long gcd3 = abs(gcd(a3, b3));
a3 /= gcd3;
b3 /= gcd3;
print(a3, b3);
}
printf("\n");
}

int main(){
long long a1, b1, a2, b2;
long long c1 = 0, c2 = 0;
scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);
//先化简到最简形式，非带分数形式
long long gcd1 = abs(gcd(a1, b1));
a1 /= gcd1;
b1 /= gcd1;
long long gcd2 = abs(gcd(a2, b2));
a2 /= gcd2;
b2 /= gcd2;
//统一用最简形式参与运算