1034. 有理数四则运算(20)-PAT乙级真题

1034.
有理数四则运算(20)

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

JAVA版请戳->>PAT
乙级 1034. 有理数四则运算(20) Java版

#include <iostream>
#include <cstdio>
using namespace std;
long long int a, b, c, d;

//辗转相除法：求最大公约数
long long int gcd(long long int t1, long long int t2) {
return t2 == 0 ? t1 : gcd(t2, t1 % t2);
}//本来自己写了个,然后超时了= =||

void func(long long int m, long long int n) {
int flag1 = 0;
int flag2 = 0;
if (n == 0) {
cout << "Inf";
return ;
}
if (m == 0) {
cout << 0;
return ;
}

if (m < 0) {
m = 0 - m;
flag1 = 1;
}
if (n < 0) {
n = 0 - n;
flag2 = 1;
}
int flag = 0;
if (flag1 == 1 && flag2 == 1) {
flag = 0;
} else if (flag1 == 1 || flag2 == 1) {
flag = 1;
}
if (m == n) {
if (flag == 1)
cout << "(-1)";
else
cout << "1";
return;
}

long long int x = m % n;
long long int y = m / n;
if (x == 0) {
if (flag == 0)
cout << y;
else
cout << "(-" << y << ")";
return ;
} else {
long long int t1 = m - y * n;
long long int t2 = n;
long long int t = gcd(t1, t2);
t1 = t1 / t;
t2 = t2 / t;
if (flag == 1) {
cout << "(-";
if (y != 0)
cout << y << " " << t1 << "/" << t2;
else
cout << t1 << "/" << t2;
cout << ")";
} else {
if (y != 0)
cout << y << " " << t1 << "/" << t2;
else
cout << t1 << "/" << t2;
}
}
}

void add() {
long long int m, n;
m = a * d + b * c;
n = b * d;
func(a, b);
cout << " + ";
func(c, d);
cout << " = ";
func(m, n);
cout << endl;
}

void min() {
long long int m, n;
m = a * d - b * c;
n = b * d;
func(a, b);
cout << " - ";
func(c, d);
cout << " = ";
func(m, n);
cout << endl;
}

void multi() {
long long int m, n;
m = a * c;
n = b * d;
func(a, b);
cout << " * ";
func(c, d);
cout << " = ";
func(m, n);
cout << endl;
}

void div() {
long long int m, n;
m = a * d;
n = b * c;
func(a, b);
cout << " / ";
func(c, d);
cout << " = ";
func(m, n);
cout << endl;
}

int main() {
scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
add();
min();
multi();
div();
return 0;
}