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。
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
#include<iostream>
#include<string>
using namespace std;
int gongyue(int a, int b)//求最大公约数
{
if (a < b)
swap(a, b);
int c;
c = a%b;
while (c != 0)
{
a = b;
b = c;
c = a%b;
}
return b;
}
void out(int a, int b)//输出格式
{
if (a == 0)
{
cout << 0;
return;
}
if (b == 0)
{
cout << "Inf";
return;
}
int flag = 0;//表示正数
if (double(a) / double(b) < 0)
flag = 1;
int c = abs(gongyue(a, b));
a = abs(a / c);
b = abs(b / c);
int k = a/b;
a = a%b;
if (flag == 0)//正数
{
if (k != 0)
{
cout << k;
if (a == 0)
return;
cout << " ";
}
cout << a << "/" << b;
}
else if (flag == 1)//负数
{
cout << "(-";
if (k != 0)
{
cout << k;
if (a == 0)
{
cout << ")";
return;
}
cout << " ";
}
cout << a << "/" << b << ")";
}
}
int main()
{
int a1, b1, a2, b2;
scanf("%d/%d", &a1, &b1);
scanf("%d/%d", &a2, &b2);
//cout << a1 << b1 << a2 << b2;
int gongbei = b1*b2 / gongyue(b1, b2);//最小公倍数
//cout << gongbei<<endl;
int a3, b3;
//加法
b3 = gongbei;
a3 = b3 / b1*a1 + b3 / b2*a2;
out(a1, b1);
cout << " + ";
out(a2, b2);
cout << " = ";
out(a3, b3);
cout << endl;
//减法
a3 = b3 / b1*a1 - b3 / b2*a2;
out(a1, b1);
cout << " - ";
out(a2, b2);
cout << " = ";
out(a3, b3);
cout << endl;
//乘法
a3 = a1*a2;
b3 = b1*b2;
out(a1, b1);
cout << " * ";
out(a2, b2);
cout << " = ";
out(a3, b3);
cout << endl;
//除法
a3 = a1*b2;
b3 = a2*b1;
out(a1, b1);
cout << " / ";
out(a2, b2);
cout << " = ";
if (b3 == 0)
cout << "Inf";
else
out(a3, b3);
cout << endl;
system("pause");
return 0;
}