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
思路
简单来说就是分数的加减乘除的问题
- 分数化简的思路——求分母和分子的最大公约数，同时÷最大公约数
- 分数相加的思路——求两个分母的最大公倍数，分子乘以对应的因子，相加即可
- 设(a,b)的最大公约数为(d)，有性质：(最大公倍数a*b/d = 最大公倍数)
- 乘除法注意判断是否为0即可
- 为了避免溢出问题，我这里统一采用了
long long型
有更好的写法是用结构体写，这样代码看起来会更好看
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b)
{
if(b == 0) return a;
else return gcd(b, a%b);
} //求最大公约数
void reduction(ll &a1, ll &b1)
{
if(a1 == 0)
return;
ll t = gcd(a1, b1);
a1 /= t;
b1 /= t;
} //分子分母约分
void standard_print(ll up, ll down)
{
if(up * down < 0) //保证负数的时候负数出现在分子
{
up = -abs(up);
down = abs(down);
}
if(up == 0)
{
printf("0");
return;
}
if(abs(up) >= abs(down))
{
ll k = up / down;
up -= k * down;
up = abs(up);
if(k < 0)
if(up == 0)
printf("(-%lld)", abs(k));
else printf("(-%lld %lld/%lld)", abs(k), up, down);
else
if(up == 0)
printf("%lld", k);
else printf("%lld %lld/%lld", k, up, down);
}else
{
if(up < 0)
printf("(%lld/%lld)", up, down);
else
printf("%lld/%lld", up, down);
}
} //控制标准输出形式，按照题目要求
void do_add(ll a1, ll b1, ll a2, ll b2)
{
ll gcd_value = gcd(b1, b2);
ll lcm_value = b1 / gcd_value * b2;
ll up = a1 * (lcm_value / b1) + a2 * (lcm_value / b2);
ll down = lcm_value;
reduction(up, down);
standard_print(a1, b1);
printf(" + ");
standard_print(a2, b2);
printf(" = ");
standard_print(up, down);
printf("
");
}
void do_sub(ll a1, ll b1, ll a2, ll b2)
{
ll gcd_value = gcd(b1, b2);
ll lcm_value = b1 / gcd_value * b2;
ll up = a1 * (lcm_value / b1) - a2 * (lcm_value / b2); //加法和减法唯一的区别就在这
ll down = lcm_value;
reduction(up, down);
standard_print(a1, b1);
printf(" - ");
standard_print(a2, b2);
printf(" = ");
standard_print(up, down);
printf("
");
}
void do_mult(ll a1, ll b1, ll a2, ll b2)
{
ll up, down;
if(a1 == 0 || a2 == 0) //任意一乘数为0则整体为0，不用再计算了
{
up = 0;
}else
{
up = a1 * a2;
down = b1 * b2;
reduction(up, down);
}
standard_print(a1, b1);
printf(" * ");
standard_print(a2, b2);
printf(" = ");
standard_print(up, down);
printf("
");
}
void do_ext(ll a1, ll b1, ll a2, ll b2)
{
ll up, down;
bool illegal = false; //判断除法是否合法
if(a2 == 0)
{
illegal = true;
}else
{
up = a1 * b2;
down = b1 * a2;
reduction(up, down);
}
standard_print(a1, b1);
printf(" / ");
standard_print(a2, b2);
printf(" = ");
if(illegal)
printf("Inf");
else standard_print(up, down);
}
int main()
{
ll a1,b1,a2,b2;
scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);
reduction(a1, b1); reduction(a2, b2);
// cin >> a1 >> b1;
// standard_print(a1, b1);
do_add(a1, b1, a2, b2);
do_sub(a1, b1, a2, b2);
do_mult(a1, b1, a2, b2);
do_ext(a1, b1, a2, b2);
return 0;
}
引用
https://pintia.cn/problem-sets/994805260223102976/problems/994805287624491008