C++ 如何化简分数
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How to simplify a fraction
提问by fuddin
I want to simplify a fraction in my application. The fraction is like, x/y where x and y are integers. I want to simplify the fraction to its simplest form. Can anyone please give me hints how to do it. Thanks in advance.
我想简化我的应用程序中的一小部分。分数类似于 x/y，其中 x 和 y 是整数。我想将分数简化为最简单的形式。任何人都可以给我提示如何做到这一点。提前致谢。
回答by cnicutar
- Compute the greatest common divisor for x and y
- Divide both of them by the GCD
- 计算 x 和 y 的最大公约数
- 将两者除以 GCD
Euclid's algorithmis an easy way to compute the GCD.
Euclid 算法是计算 GCD 的一种简单方法。
回答by Peter Alexander
Divide both by gcd(x,y)
除以 gcd(x,y)
The Binary GCD algorithmis a fast way to compute the GCD on a computer.
该二进制GCD算法是计算在计算机上的GCD的快捷方式。
回答by Zaryab khan
#include<iostream>
using namespace std;
struct fraction
{
int n1, d1, n2, d2, s1, s2;
};
void simplification(int a,int b)
{
bool e = true;
int t; int z;
for (int i = (a*b); i > 1;i--)
{ if ((a%i==0)&&(b%i==0))
{
t = a / i;
z = b / i;
}
else
{
e = false;
}
}
cout << "simplest form=" << t << "/" << z << endl;
}
void sum(int num1, int deno1, int num2, int deno2)
{
int k,y;
k = num1* deno2 + num2*deno1;
y = deno2*deno1;
cout << "addition of given fraction = " << k << "/" << y << endl;
simplification(k, y);
}
void sub(int num1, int deno1, int num2, int deno2)
{
int k, y;
k = num1*deno2 - num2*deno1;
y = deno1*deno2;
cout << "Substraction of given fraction = " << k << "/" << y << endl;
}
void mul(int num1, int deno1, int num2, int deno2)
{
int k, y;
k = num1*num2;
y = deno1*deno2;
cout << "multiplication of given fration= " << k<< "/" <<y; cout<< endl;
simplification(k, y);
}
void div(int num1, int deno1, int num2, int deno2)
{
int k, y;
;
k = num1*deno1;
y = deno1*num2;
cout << "division of given fraction" << k << "/" << y << endl;
simplification(k, y);
}
int main()
{ fraction a;
cout << "enter numirator of f1=";cin >> a.n1;
cout << "enter denominator of f1=";cin >> a.d1;
cout << "enter numirator of f2=";cin >> a.n2;
cout << "enter denominator of f2=";cin >> a.d2;
cout << "f1= " << a.n1 << "/" << a.d1 << endl;
cout << "f2= " << a.n2 << "/" << a.d2 << endl;
mul(a.n1, a.d1, a.n2, a.d2);
div(a.n1, a.d1, a.n2, a.d2);
sub(a.n1, a.d1, a.n2, a.d2);
sum(a.n1, a.d1, a.n2, a.d2);
system("pause");
}
