如何简化分数

计算最伟大的普通分频器 X 和 y

将它们分开 GCD

http://en.wikipedia.org/wiki/G ... rithm
- 这是一种计算的简单方法 GCD.

除了两者
http://en.wikipedia.org/wiki/G ... visor
http://en.wikipedia.org/wiki/Binary_GCD_algorithm
- 这是一种快速计算的方法 GCD 在电脑上面。

[code]#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 &gt; 1;i--/
{ if //a%i==0/&amp;&amp;/b%i==0//
{
t = a / i;
z = b / i;

}
else
{
e = false;
}
}
cout &lt;&lt; "simplest form=" &lt;&lt; t &lt;&lt; "/" &lt;&lt; z &lt;&lt; endl;

}
void sum/int num1, int deno1, int num2, int deno2/
{
int k,y;
k = num1* deno2 + num2*deno1;
y = deno2*deno1;
cout &lt;&lt; "addition of given fraction = " &lt;&lt; k &lt;&lt; "/" &lt;&lt; y &lt;&lt; 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 &lt;&lt; "Substraction of given fraction = " &lt;&lt; k &lt;&lt; "/" &lt;&lt; y &lt;&lt; endl;

}
void mul/int num1, int deno1, int num2, int deno2/
{
int k, y;

k = num1*num2;
y = deno1*deno2;
cout &lt;&lt; "multiplication of given fration= " &lt;&lt; k&lt;&lt; "/" &lt;<y; "="" "division="" "enter="" "f2=" &lt;&lt; a.n2 &lt;&lt; " "pause"="" ;="" <="" <<="" a.d1="" a.d1,="" a.d2="" a.n1,="" a.n2,="" a;="" code]="" cout="" cout<<="" deno1,="" deno2="" denominator="" div="" endl;="" enter="" f1=" &lt;&lt; a.n1 &lt;&lt; " f2=";cin &gt;&gt; a.d2;
cout &lt;&lt; " fraction="" fraction"="" given="" int="" k="" k,="" main="" mul="" num1,="" num2,="" numirator="" of="" simplification="" sub="" sum="" system="" void="" y="" y;="" {="" }="" }[="">
</y;></iostream>