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如何将小数转换为分数?

[英]How to convert decimal to fractions?

What I need to convert decimal to fractions.我需要将十进制转换为分数。 It is easy to convert to 10's feet.很容易转换为 10 英尺。

1.5 => 15/10

This can do via this code:这可以通过以下代码完成:

public class Rational {

    private int num, denom;

    public Rational(double d) {
        String s = String.valueOf(d);
        int digitsDec = s.length() - 1 - s.indexOf('.');
        int denom = 1;
        for (int i = 0; i < digitsDec; i++) {
            d *= 10;    
            denom *= 10;
        }

        int num = (int) Math.round(d);
        this.num = num;
        this.denom = denom;
    }

    public Rational(int num, int denom) {
        this.num = num;
        this.denom = denom;
    }

    public String toString() {
        return String.valueOf(num) + "/" + String.valueOf(denom);
    }

    public static void main(String[] args) {
        System.out.println(new Rational(1.5));
    }
}

But what I want is但我想要的是

1.5 => 3/2

and I don't get how to proceed.我不知道如何继续。 My question is not a duplication.我的问题不是重复。 Because other related question is C#.因为其他相关问题是 C#。 This is java.这是爪哇。

static private String convertDecimalToFraction(double x){
    if (x < 0){
        return "-" + convertDecimalToFraction(-x);
    }
    double tolerance = 1.0E-6;
    double h1=1; double h2=0;
    double k1=0; double k2=1;
    double b = x;
    do {
        double a = Math.floor(b);
        double aux = h1; h1 = a*h1+h2; h2 = aux;
        aux = k1; k1 = a*k1+k2; k2 = aux;
        b = 1/(b-a);
    } while (Math.abs(x-h1/k1) > x*tolerance);

    return h1+"/"+k1;
}

I got this answer from here .我从这里得到了这个答案。 All I had to do is convert his answer to java.我所要做的就是将他的答案转换为 java。

You should find the greatest common divisor of the resulted numbers and divide the numerator and denominator by it.您应该找到结果数的最大公约数,并将分子和分母除以它。

Here is one way to do it:这是一种方法:

public class Rational {

    private int num, denom;

    public Rational(double d) {
        String s = String.valueOf(d);
        int digitsDec = s.length() - 1 - s.indexOf('.');
        int denom = 1;
        for (int i = 0; i < digitsDec; i++) {
            d *= 10;    
            denom *= 10;
        }

        int num = (int) Math.round(d);
        int g = gcd(num, denom);
        this.num = num / g;
        this.denom = denom /g;
    }

    public Rational(int num, int denom) {
        this.num = num;
        this.denom = denom;
    }

    public String toString() {
        return String.valueOf(num) + "/" + String.valueOf(denom);
    }

    public static int gcd(int num, int denom) {
          ....
    }

    public static void main(String[] args) {
        System.out.println(new Rational(1.5));
    }
}

Given double x >= 0, int p, int q, find p/q as closest approximation:给定 double x >= 0, int p, int q,找到 p/q 作为最接近的近似值:

  • iterate on q from 1 upwards, determine p above and below;从 1 向上迭代 q,确定上下 p; check deviations检查偏差

So (not tested):所以(未测试):

public static Rational toFraction(double x) {
    // Approximate x with p/q.
    final double eps = 0.000_001;
    int pfound = (int) Math.round(x);
    int qfound = 1;
    double errorfound = Math.abs(x - pfound);
    for (int q = 2; q < 100 && error > eps; ++q) {
        int p = (int) (x * q);
        for (int i = 0; i < 2; ++i) { // below and above x
            double error = Math.abs(x - ((double) p / q));
            if (error < errorfound) {
                pfound = p;
                qfound = q;
                errorfound = error;
            }
            ++p;
        }
    }
    return new Rational(pfound, qfound);
}

You could try it for Math.PI and E.你可以试试 Math.PI 和 E。

Here is a simple algorythme :这是一个简单的算法:

numerato = 1.5
denominator = 1;

while (!isInterger(numerator*denominator))
do
    denominator++;
done

return numerator*denominator + '/' + denominator


// => 3/2

You just have to implement it in java + implement the isInteger(i) where i is a float .你只需要在java中实现它+实现isInteger(i) ,其中i是一个float

Including the method to find highest common factor and modifying toString method, solves your question i suppose.包括找到最高公因数的方法和修改 toString 方法,我想解决了你的问题。

public String toString() {
        int hcf = findHighestCommonFactor(num, denom);
        return (String.valueOf(num/hcf) + "/" + String.valueOf(denom/hcf));

    }

    private int findHighestCommonFactor(int num, int denom) {
        if (denom == 0) {
            return num;
        }
        return findHighestCommonFactor(denom, num % denom);
    }

Not only for the decimal number 1.5 , for all you can use the following steps:不仅对于十进制数1.5 ,对于所有您可以使用以下步骤:

  1. Find Number of decimal digits:查找小数位数:

    double d = 1.5050;//Example I used

    double d1 = 1;

    String text = Double.toString(Math.abs(d));

    int integerPlaces = text.indexOf('.');

    int decimalPlaces = text.length() - integerPlaces - 1;

    System.out.println(decimalPlaces);//4

  2. Then convert to integer:然后转换为整数:

    static int ipower(int base, int exp) {

     int result = 1; for (int i = 1; i <= exp; i++) { result *= base; } return result; }

    //using the method

    int i = (int) (d*ipower(10, decimalPlaces));

    int i1 = (int) (d1*ipower(10, decimalPlaces));

    System.out.println("i=" + i + " i1 =" +i1);//i=1505 i1 =1000

  3. Then find highest common factor然后求最大公因数

    private static int commonFactor(int num, int divisor) {

     if (divisor == 0) { return num; } return commonFactor(divisor, num % divisor); }

//using common factor

int commonfactor = commonFactor(i, i1);

System.out.println(commonfactor);//5

  1. Finally print results:最后打印结果:

    System.out.println(i/commonfactor + "/" + i1/commonfactor);//301/200

Here you can find:您可以在这里找到:

  public static void main(String[] args) {

        double d = 1.5050;
        double d1 = 1;

        String text = Double.toString(Math.abs(d));
        int integerPlaces = text.indexOf('.');
        int decimalPlaces = text.length() - integerPlaces - 1;

        System.out.println(decimalPlaces);
        System.out.println(ipower(10, decimalPlaces));

        int i = (int) (d*ipower(10, decimalPlaces));
        int i1 = (int) (d1*ipower(10, decimalPlaces));      

        System.out.println("i=" + i + " i1 =" +i1);

        int commonfactor = commonFactor(i, i1);
        System.out.println(commonfactor);

        System.out.println(i/commonfactor + "/" + i1/commonfactor);


    }

    static int ipower(int base, int exp) {
        int result = 1;
        for (int i = 1; i <= exp; i++) {
            result *= base;           
        }

        return result;
    }

    private static int commonFactor(int num, int divisor) {
        if (divisor == 0) {
            return num;
        }
        return commonFactor(divisor, num % divisor);
    }

I tried adding this as an edit, but it was denied.我尝试将其添加为编辑,但被拒绝。 This answer builds off of @Hristo93's answer but finishes the gcd method:这个答案建立客@ Hristo93的答案,但完成GCD的方法:

public class DecimalToFraction {

    private int numerator, denominator;

    public Rational(double decimal) {
        String string = String.valueOf(decimal);
        int digitsDec = string.length() - 1 - s.indexOf('.');
        int denominator = 1; 

        for (int i = 0; i < digitsDec; i++) {
            decimal *= 10;    
            denominator *= 10;
        }

        int numerator = (int) Math.round(decimal);
        int gcd = gcd(numerator, denominator); 

        this.numerator = numerator / gcd;
        this.denominator = denominator /gcd;
    }

    public static int gcd(int numerator, int denom) {
        return denominator == 0 ? numerator : gcm(denominator, numerator % denominator);
    }

    public String toString() {
        return String.valueOf(numerator) + "/" + String.valueOf(denominator);
    }

    public static void main(String[] args) {
        System.out.println(new Rational(1.5));
    }
}

I prepared a solution for this question.我为这个问题准备了一个解决方案。 Maybe it's look like primitive but working.也许它看起来像原始但有效。 I tested many decimal number.我测试了很多十进制数。 At least it can converting 1.5 to 3/2 :)至少它可以将 1.5 转换为 3/2 :)

public String kesirliYap(Double sayi){
    String[] a=payPaydaVer(sayi);
    return a[0]+"/"+a[1];
}
public String[] payPaydaVer(Double sayi){
long pay;
long payda;

  DecimalFormat df=new DecimalFormat("#");
    df.setRoundingMode(RoundingMode.FLOOR);
    String metin=sayi.toString();        
    int virguldenSonra=(metin.length() -metin.indexOf("."))-1;
    double payyda=Math.pow(10,virguldenSonra);
    double payy=payyda*sayi;
    String pays=df.format(payy);
    String paydas=df.format(payyda);
    pay=Long.valueOf(pays);
    payda=Long.valueOf(paydas);


   String[] kesir=sadelestir(pay,payda).split(",");

   return kesir;
}

private String sadelestir(Long pay,Long payda){
    DecimalFormat df=new DecimalFormat("#");
    df.setRoundingMode(RoundingMode.FLOOR);
    Long a=pay<payda ? pay : payda;
    String b = "",c = "";
    int sayac=0;
    for(double i = a;i>1;i--){
      double payy=pay/i;
      double paydaa=payda/i;
      String spay=df.format(payy);
      String spayda=df.format(paydaa);
      Long lpay=Long.valueOf(spay);
      Long lpayda=Long.valueOf(spayda);
      if((payy-lpay)==0&&(paydaa-lpayda)==0){
          b=df.format(pay/i);
          c=df.format(payda/i);
          sayac++;
          break;
      }

    }

    return sayac>0 ?  b+","+c:pay+","+payda;
}

First of all, if you want to convert a decimal number, you need to know the state of the situation before you convert it, let us say you have 0.333333, number 3 is being repeated infinitely.首先,如果你想转换一个十进制数,你需要在转换之前知道情况的状态,假设你有0.333333,数字3被无限重复。 We all know that 0.333333 is 1/3 .我们都知道 0.333333 是 1/3 。 Some people think that multiplying by the number of digits after the decimal point will convert it.有人认为乘以小数点后的位数就可以换算了。 That is in some cases is false, and the other is true.那是在某些情况下是错误的,而另一种情况是正确的。 It is something related to mathematics.这是与数学有关的东西。 Another situation is 0.25, take the numbers after the decimal point and divide them by 100 and simplify them, which equals to 1/4.另一种情况是0.25,取小数点后的数除以100并化简,等于1/4。 states have been covered, one more to go, but I am not going to explain it because it is long.状态已被覆盖,还有一个,但我不打算解释它,因为它很长。

However, in mathematics we have 3 states for converting a decimal number into a fraction, I am not going to explain them because It will take a lot of space and time, I have already written a program for this problem.但是,在数学中,我们有 3 种状态将十进制数转换为分数,我不打算解释它们,因为这会占用大量空间和时间,我已经为这个问题编写了一个程序。 This is the code:这是代码:

import java.math.BigDecimal;
import java.math.BigInteger;

public class Main {
    static BigDecimal finalResult = new BigDecimal("0");
    
    static boolean check(short[] checks) {
        boolean isContinues = true;
        int index = -1;
        for (short ind : checks) {
            index++;
            if (ind==1) {
                
            }
            else if (ind==0) {
                isContinues = false;
                break;
            }
            else if (ind==-1) {
                if (index==0) {
                    isContinues = false;
                }
                break;
            }
        }
        
        return isContinues;
    }
    static int[] analyzeDecimal() { // will return int[3]
        int[] analysis = new int[3];
        int dot = finalResult.toString().indexOf(".");
        String num = finalResult.toString();
        int state = -1;
        int firstPart = 0; // first part will be compared with each secondPart!
        int secondPart = 0; 
        String part = ""; // without the dot
        int index = 0; // index for every loop!
        int loop = 6;
        int originalLoop = loop;
        int size = 0; // until six!
        int ps = -1;
        short[] checks = new short[] {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}; // 10 compares for each part!
        // length of checks is 10!
        int continues = -1; // -1 means there is no continues part!
        boolean stop = false;
        while (true) { // while for size!
            if (size!=6) {
            while (true) { // we need to compare a part with a part!
                // while for loop
                // 6 loops, every loop will increase the compared part by 1!
                if (loop!=-1) { // TODO : check every part with the increasing pos
                    firstPart = dot+1+(originalLoop-loop); // changed
                    try {
                        part = num.substring(firstPart, firstPart+(size+1));
                    }
                    catch (StringIndexOutOfBoundsException ex) {
                        break;
                    }
                    int partSize = part.length();
                    int afterDecimal = num.length()-(dot+1);
                    while (index!=checks.length && 
                        firstPart+partSize+index*partSize-(dot+1)<=afterDecimal) { // while for index!
                        secondPart = firstPart+partSize+index*partSize;
                        String comparedPart;
                        try {
                            comparedPart = num.substring(secondPart, secondPart+partSize);
                        }
                        catch (StringIndexOutOfBoundsException ex) {
                            break;
                        }
                        if (part.equals(comparedPart)) {
                            checks[index] = 1;
                        }
                        else {
                            checks[index] = 0;
                        }
                        index++;
                    }
                    index = 0;
                    if (check(checks)) {
                        stop = true;
                        continues = firstPart;
                        ps = partSize;
                    }
                    for (int i = 0 ; i!=10 ; i++) {
                        checks[i] = -1;
                    }
                }
                else { // finished!
                    break;
                }
                loop--;
                if (stop) {
                    break;
                }
            }
            loop = originalLoop;
            size++;
            if (stop) {
                break;
            }
            }
            else {
                break;
            }
        }
        if (continues==-1) {
            state = 2;
        }
        else {
            if (dot+1==continues) {
                state = 1;
            }
            else {
                state = 0;
            }
        }
        analysis[0] = state;
        analysis[1] = continues;
        analysis[2] = ps;
        
        return analysis;
    }
    static String convertToStandard() {
        // determine the state first : 
        int[] analysis = analyzeDecimal();
        int dot = finalResult.toString().indexOf('.')+1;
        int continues = analysis[1];
        int partSize = analysis[2]; // how many steps after the continues part
        if (analysis[0]==0) { // constant + continues
            String number = finalResult.toString().substring(0, continues+partSize);
            int numOfConst = continues-dot;
            int numOfDecimals = continues+partSize-dot;
            int den = (int)(Math.pow(10, numOfDecimals)-Math.pow(10, numOfConst)); // (10^numOfDecimals)-(10^numOfConst);
            int num;
            int toSubtract = Integer.parseInt(number.substring(0, dot-1)+number.substring(dot, dot+numOfConst));
            if (number.charAt(0)==0) {
                num = Integer.parseInt(number.substring(dot));
            }
            else {
                num = Integer.parseInt(number.replace(".", ""));
            }
            num -= toSubtract;
            return simplify(num, den);
        }
        
        else if (analysis[0]==1) { // continues 
            int num, den;
            // we always have  to subtract by only one x!
            String n = finalResult.toString().substring(0, dot+partSize).replace(".", "");
            num = Integer.parseInt(n);
            den = nines(partSize);
            int toSubtract = Integer.parseInt(finalResult.toString().substring(0, dot-1));
            num -= toSubtract;
            return simplify(num, den);
        }
        else if (analysis[0]==2) { // constant
            partSize = finalResult.toString().length()-dot;
            int num = Integer.parseInt(finalResult.toString().replace(".", ""));
            int den = (int)Math.pow(10, partSize);
            return simplify(num, den);
        }
        else {
            System.out.println("[Error] State is not determined!");
        }
        
        return "STATE NOT DETERMINED!";
    }
    static String simplify(int num, int den) {
        BigInteger n1 = new BigInteger(Integer.toString(num));
        BigInteger n2 = new BigInteger(Integer.toString(den));
        BigInteger GCD = n1.gcd(n2);
        String number = Integer.toString(num/GCD.intValue())+"/"+Integer.toString(den/GCD.intValue());
        
        return number;
    }
    static int nines(int n) {
        StringBuilder result = new StringBuilder();
        while (n!=0) {
            n--;
            result.append("9");
        }
        return Integer.parseInt(result.toString());
    }
    public static void main(String[] args) {
        finalResult = new BigDecimal("1.222222");
        System.out.println(convertToStandard());
    }
}

The program above will give you an optimal result with high precision.上面的程序将为您提供高精度的最佳结果。 All you have to do is change the finalResult variable in the main function.您所要做的就是更改主函数中的 finalResult 变量。

Well check this simple implementation, I have not used any GCD or something, instead, I have put the logic for the numerator and keep on incrementing till the logic is not fulfilled.好吧,检查一下这个简单的实现,我没有使用任何 GCD 或其他东西,相反,我已经为分子添加了逻辑并继续递增,直到逻辑不满足为止。

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    
    System.out.println("Enter the decimal number:");
    double d = scan.nextDouble();
    
    int denom = 1;
    boolean b = true;
    while(b) {
        String[] s = String.valueOf(d * denom).split("\\.");
        if(s[0].equals(String.valueOf((int)(d * denom))) && s[1].equals("0")) {
            break;
        }
        denom++;
    }
    
    if(denom == 1) {
        System.out.println("Input a decimal number");
    }
    else {
        System.out.print("Fraction: ");
        System.out.print((int)(d*denom)+"/"+denom);
    }
}

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