mirror of
https://github.com/TheAlgorithms/Java.git
synced 2025-07-14 01:16:07 +08:00
@ -1,18 +1,13 @@
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import java.lang.StringBuilder;
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import java.util.*;
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import java.util.Scanner;
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import javax.swing.*;
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public class HexaDecimalToBinary {
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private final int LONG_BITS = 8;
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public void convert(String numHex) {
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//String a HexaDecimal:
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// String a HexaDecimal:
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int conHex = Integer.parseInt(numHex, 16);
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//Hex a Binary:
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// Hex a Binary:
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String binary = Integer.toBinaryString(conHex);
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//Presentation:
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// Presentation:
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System.out.println(numHex + " = " + completeDigits(binary));
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}
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@ -1,33 +1,32 @@
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import java.io.*;
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import java.util.*;
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import java.text.*;
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import java.math.*;
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import java.util.regex.*;
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import java.util.Scanner;
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public class RootPrecision {
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public static void main(String[] args) {
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//take input
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// take input
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Scanner scn = new Scanner(System.in);
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int N = scn.nextInt(); //N is the input number
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int P = scn.nextInt(); //P is precision value for eg - P is 3 in 2.564 and 5 in 3.80870.
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// N is the input number
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int N = scn.nextInt();
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// P is precision value for eg - P is 3 in 2.564 and 5 in 3.80870.
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int P = scn.nextInt();
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System.out.println(squareRoot(N, P));
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}
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public static double squareRoot(int N, int P) {
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double rv = 0; //rv means return value
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// rv means return value
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double rv;
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double root = Math.pow(N, 0.5);
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//calculate precision to power of 10 and then multiply it with root value.
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// calculate precision to power of 10 and then multiply it with root value.
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int precision = (int) Math.pow(10, P);
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root = root * precision;
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/*typecast it into integer then divide by precision and again typecast into double
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so as to have decimal points upto P precision */
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rv = (int)root;
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return (double)rv/precision;
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rv = (int) root;
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return rv / precision;
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}
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}
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