我正在尝试将十进制数转换为其分数。小数点后最多有 4 位数字。例如：- 12.34 = 1234/100 12.3456 = 123456/10000
我的代码：-
#include <stdio.h>
int main(void) {
double a=12.34;
int c=10000;
double b=(a-floor(a))*c;
int d=(int)floor(a)*c+(int)b;
while(1) {
if(d%10==0) {
d=d/10;
c=c/10;
}
else break;
}
printf("%d/%d",d,c);
return 0;
}
但我没有得到正确的输出，十进制数只有双精度。请指导我应该做什么。
如果您的浮点数是x，那么超过 10000 的分数的分子将是 的整数部分(x + 0.00005) * 10000。是否要将分数简化为最简单的项（即除以分子和分母的 gcd）取决于您。
#include <stdio.h>
int main(void) {
double a = 12.34;
int c = 10000;
double b = (a - floor(a)) * c;
int d = (int)floor(a) * c + (int)(b + .5f);
printf("%f %d\n", b, d);
while(1) {
if(d % 10 == 0) {
d = d / 10;
c = c / 10;
}
else break;
}
printf("%d/%d\n", d, c);
return 0;
}
问题是b得到了 3400.00，但是当你(int) b得到 3399 时，你需要添加0.5以便数字可以截断为 3400。
获得 3400.00 与拥有 3400 不同，3400.00 意味着该数字四舍五入为 3400，这就是为什么当您执行 (int) 3400.00 时，它假定最接近的整数（小于您要转换的数字）是 3399，但是，当您添加0.5 到那个数字，最后一个最接近的整数现在是 3400。
如果您想更深入地了解浮点运算，请阅读每个计算机科学家应该了解的关于浮点运算的知识
我的解决方案很简单，“懒惰”，通过迭代运行，没什么花哨的。
在大多数拥有不错数学库的语言中，您只需要算法本身即可。
但是在 bc 中，您需要实现简单的功能，例如
int() to return integer part of a number ,
abs() to return absolute value ,
float() to return floating part of a number ,
round() to round to nearest integer.
如果在 (1/eps) 次迭代后没有找到任何东西，则循环以最后一个结果中断。
eps=10^-4 /*Tweak for more or less accuracy */
define int(x) {
auto s ;
s = scale ;
scale = 0 ;
x /= 1 ;
scale = s ;
return x ;
}
define round(x) { return int(x+.5-(x<0)) ; }
define abs(x) { if ( x < 0 ) x=-x ; return x ; }
define float(x) { return abs(x-int(x)) ; }
define void frac(x) {
auto f, j, n, z ;
f = float(x) ;
j = 1 / eps ;
z = .5 ;
if ( f != 0 ) {
while ( ( n++ < j ) && ( abs( z - round(z) ) > eps ) ) z = n / f ;
n -= 1 ;
if ( x < 0 ) n = -n ;
x = int(x)
z = round(z) ;
print n + x*z , "/" , z , " = "
if ( x != 0 ) print x , " + " , n , "/" , z , " = "
}
print x+n/z , "\n" ;
}
使用标准精度（eps=.0001），您可以得到：
frac(-.714285)
-5/7 = -.71428571428571428571
sqrt(2)
1.414213562373
frac(sqrt(2))
19601/13860 = 1 + 5741/13860 = 1.414213564213
6-7/pi
3.77183080
eps=.000001 ; frac(6-7/pi)
1314434/348487 = 3 + 268973/348487 = 3.77183080
这是我使用的算法。这是一个迭代过程，其工作原理如下：
这种方法的一些特点是：
我在 github 上发布了这个算法的代码——https: //github.com/tnbezue/fraction
这是个有趣的问题。我认为您最好从阅读计算“最大公约数”的倍数方法开始（http://en.wikipedia.org/wiki/Greatest_common_divisor是一个很好的来源）。
实现一个快速而肮脏的算法，像用笔和纸一样进行这些计算，然后研究双精度数的表示方式（符号、指数、尾数）并改进您的算法以利用这种表示。
可悲的是，如果不编写您的代码，我将无能为力。
用 c++ 创建的一种算法，可以将小数转换为分数。
#include <iostream>
using namespace std;
// converts the string half of the inputed decimal number into numerical values
void converting (string decimalNumber, float& numerator, float& denominator )
{
float number;
string valueAfterPoint = decimalNumber.substr(decimalNumber.find(".") + 1,((decimalNumber.length() -1) )); // store the value after the decimal into a valueAfterPoint
cout << valueAfterPoint<< " "<< endl;
int length = valueAfterPoint.length(); //stores the length of the value after the decimal point into length
numerator = atof(valueAfterPoint.c_str()); // converts the string type decimal number into a float value and stores it into the numerator
// loop increases the decimal value of the numerator and the value of denominator by multiples of ten as long as the length is above zero of the decimal
cout << length<< endl;
for (; length > 0; length--)
{
numerator *= 10;
}
do
denominator *=10;
while (denominator < numerator);
}
// simplifies the the converted values of the numerator and denominator into simpler values for an easier to read output
void simplifying (float& numerator, float& denominator)
{
int maximumNumber = 9; //Numbers in the tenths place can only range from zero to nine so the maximum number for a position in a poisitino for the decimal number will be nine
bool isDivisble; // is used as a checker to verify whether the value of the numerator has the found the dividing number that will a value of zero
// Will check to see if the numerator divided denominator is will equal to zero
if(int(numerator) % int(denominator) == 0)
{
numerator /= denominator;
denominator = 1;
return;
}
//check to see if the maximum number is greater than the denominator to simplify to lowest form
while (maximumNumber < denominator)
{
maximumNumber *=10;
}
// the maximum number loops from nine to zero. This conditions stops if the function isDivisible is true
for(; maximumNumber > 0; maximumNumber --)
{
isDivisble = ((int(numerator) % maximumNumber == 0) && int(denominator)% maximumNumber == 0);
cout << numerator << denominator <<" " <<endl;
if(isDivisble)
{
numerator /= maximumNumber; // when is divisible true numerator be devided by the max number value for example 25/5 = numerator = 5
denominator /= maximumNumber; //// when is divisible true denominator be devided by the max number value for example 100/5 = denominator = 20
}
// stop value if numerator and denominator is lower than 17 than it is at the lowest value
int stop = numerator + denominator;
if (stop < 17)
{
return;
}
}
}
int main()
{
string decimalNumber;
float numerator = 0;
float denominator = 1;
cout << "Enter the decimal number";
cin >> decimalNumber;
//convert function
converting(decimalNumber, numerator, denominator);
//call simplyfication funcition
simplifying(numerator, denominator);
cout<< "Fraction: "<< numerator << "/" << denominator<< endl;
return 0;
}