[PAT B1034/A1088]Rational Arithmetic
本题是PAT乙级1034和甲级1088题，这里先放英文原题，要做乙级题的小伙伴们可以跳过直接看后面的中文原题
题目描述
1088 Rational Arithmetic （20 分)For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.
输入格式
Each input file contains one test case, which gives in one line the two rational numbers in the format a1/b1 a2/b2. The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.
输出格式
For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is number1 operator number2 = result. Notice that all the rational numbers must be in their simplest form k a/b, where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output Inf as the result. It is guaranteed that all the output integers are in the range of long int.
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
--------------------------中文原题---------------------------
题目描述
1034 有理数四则运算 （20 分)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。
解析
题目大意就是输入两个数，以上面的格式输出他们的和差积商，需要注意的地方就是如果分母为0，就输出Inf,详细解析可以移步至，里面有详细的注释
工欲善其事必先利其器——盘点PAT中非常好用的工具（持续更新）
#include<iostream>
using namespace std;
struct Fraction
{
long long up, down;
};
long long gcd(long long a,long long b)
{
return !b ? a : gcd(b, a%b);
}
Fraction reduction(Fraction f)
{
if (f.down < 0) {
f.up = -f.up;f.down = -f.down;}if (f.up == 0) f.down = 1;else {
long long t = gcd(abs(f.up), abs(f.down));f.up /= t;f.down /= t;}return f;
}
Fraction add(Fraction a, Fraction b)
{
Fraction f;f.up = a.up*b.down + a.down*b.up;f.down = a.down*b.down;return reduction(f);
}
Fraction difference(Fraction a, Fraction b)
{
Fraction f;f.up = a.up*b.down - a.down*b.up;f.down = a.down*b.down;return reduction(f);
}
Fraction product(Fraction a, Fraction b)
{
Fraction f;f.up = a.up*b.up;f.down = a.down*b.down;return reduction(f);
}
Fraction quotient(Fraction a, Fraction b)
{
Fraction f;f.up = a.up*b.down;f.down = a.down*b.up;return reduction(f);
}
void show(Fraction f)
{
if (f.down == 0) printf("Inf");else if (f.up < 0) {
if (f.down == 1)printf("(%lld)", f.up);else if (abs(f.up) > f.down) printf("(%lld %lld/%lld)", f.up / f.down, abs(f.up) % f.down, f.down);else printf("(%lld/%lld)", f.up, f.down);}else {
if (f.down == 1)printf("%lld", f.up);else if (abs(f.up) > f.down) printf("%lld %lld/%lld", f.up / f.down, abs(f.up) % f.down, f.down);else printf("%lld/%lld", f.up, f.down);}
}
int main()
{
Fraction f1, f2;scanf("%lld/%lld %lld/%lld", &f1.up, &f1.down, &f2.up, &f2.down);f1=reduction(f1); f2=reduction(f2);show(f1); cout << " + "; show(f2); cout << " = "; show(add(f1, f2)); cout << endl;show(f1); cout << " - "; show(f2); cout << " = "; show(difference(f1, f2)); cout << endl;show(f1); cout << " * "; show(f2); cout << " = "; show(product(f1, f2)); cout << endl;show(f1); cout << " / "; show(f2); cout << " = "; show(quotient(f1, f2)); cout << endl;return 0;
}