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 2/32/3 * (-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 2/31 2/3 - 0 = 1 2/31 2/3 * 0 = 01 2/3 / 0 = Inf 代码：
#includeusing namespace std;long long int gcd(long long int x,long long int y) { long long int z = x % y; while(z) { x = y; y = z; z = x % y; } return y;}void out(long long int x,long long int y) { if(y < 0) { y = y * (-1); x = x * (-1); } if(x > 0) { if(y == 1) printf("%lld", x); else { if(x % y == 0) printf("%lld", x / y); else { if(x < y) printf("%lld/%lld", x / gcd(x, y), y / gcd(x, y)); else printf("%lld %lld/%lld", x / y, (x % y) / gcd(x % y, y), y / gcd(x % y, y)); } } } else if(x == 0) printf("0"); else { if(y == 1) printf("(%lld)", x); else { if(x % y == 0) printf("(%lld)", x / y); else { if(abs(x) < y) printf("(%lld/%lld)", x / abs(gcd(x, y)), y / abs(gcd(x, y))); else printf("(%lld %lld/%lld)", x / y, abs(x) % y/gcd(abs(x) % y, y), y / gcd(abs(x) % y, y)); } } }}void add(long long int a1, long long int b1, long long int a2, long long int b2) { out(a1, b1); printf(" + "); out(a2, b2); printf(" = "); long long int a3 = a1 * b2 + a2 * b1; long long int b3 = b1 * b2; out(a3, b3); printf("\n");}void subtract(long long int a1, long long int b1, long long int a2, long long int b2) { out(a1, b1); printf(" - "); out(a2, b2); printf(" = "); long long int a3 = a1 * b2 - a2 * b1; long long int b3 = b1 * b2; out(a3, b3); printf("\n");}void multiplication(long long int a1, long long int b1, long long int a2, long long int b2) { out(a1, b1); printf(" * "); out(a2, b2); printf(" = "); long long int a3 = a1 * a2; long long int b3 = b1 * b2; out(a3, b3); printf("\n");}void division(long long int a1, long long int b1, long long int a2, long long int b2) { out(a1, b1); printf(" / "); out(a2, b2); printf(" = "); if(a2 * b2 == 0) { printf("Inf"); return ; } long long int a3 = a1 * b2; long long int b3 = a2 * b1; out(a3, b3); printf("\n");}int main() { long long int a, b, c, d; scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d); add(a, b, c, d); subtract(a, b, c, d); multiplication(a, b, c, d); division(a, b, c, d); return 0;}