1034 有理數四則運算 (20 分)
本題要求編寫程序,計算 2 個有理數的和、差、積、商。
輸入格式:
輸入在一行中按照 a1/b1 a2/b2
的格式給出兩個分數形式的有理數,其中分子和分母全是整型範圍內的整數,負號只可能出現在分子前,分母不爲 0。
輸出格式:
分別在 4 行中按照 有理數1 運算符 有理數2 = 結果
的格式順序輸出 2 個有理數的和、差、積、商。注意輸出的每個有理數必須是該有理數的最簡形式 k a/b
,其中 k
是整數部分,a/b
是最簡分數部分;若爲負數,則須加括號;若除法分母爲 0,則輸出 Inf
。題目保證正確的輸出中沒有超過整型範圍的整數。
輸入樣例 1:
2/3 -4/2
輸出樣例 1:
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
輸入樣例 2:
5/3 0/6
輸出樣例 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)//求最大公約數
{
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)//輸出格式
{
if (a == 0)
{
cout << 0;
return;
}
if (b == 0)
{
cout << "Inf";
return;
}
int flag = 0;//表示正數
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)//正數
{
if (k != 0)
{
cout << k;
if (a == 0)
return;
cout << " ";
}
cout << a << "/" << b;
}
else if (flag == 1)//負數
{
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);//最小公倍數
//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;
//減法
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;
}