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Numeron voima Javassa

Tässä osiossa kirjoitamme Java-ohjelmia luvun tehon määrittämiseksi. Saadaksesi luvun potenssin, kerro luku sen eksponentilla.

Esimerkki:

Oletetaan, että kanta on 5 ja eksponentti 4. Saadaksesi luvun potenssin, kerro se itsellään neljä kertaa, eli (5 * 5 * 5 * 5 = 625).

Kuinka määrittää luvun teho?

  • Kanta ja eksponentti tulee lukea tai alustaa.
  • Ota toinen muuttuva potenssi ja aseta se arvoon 1 tallentaaksesi tuloksen.
  • Kerro perusarvo teholla ja tallenna tulos tehoon for- tai while-silmukan avulla.
  • Toista vaihe 3, kunnes eksponentti on nolla.
  • Tulosta tulos.

Menetelmät luvun voiman löytämiseksi

On olemassa useita menetelmiä numeron tehon määrittämiseen:

satunnainen c
  1. Java for Loop käyttö
  2. Javan käyttö silmukan aikana
  3. Rekursion käyttö
  4. Käyttämällä Math.pow()-menetelmää
  5. Bittimanipuloinnin käyttö

1. Java for Loopin käyttö

For-silmukkaa voidaan käyttää luvun tehon laskemiseen kertomalla kanta itsellään toistuvasti.

PowerOfNumber1.java

 public class PowerOfNumber1 { public static void main(String[] args) { int base = 2; int exponent = 3; int result = 1; for (int i = 0; i <exponent; i++) { result *="base;" } system.out.println(base + ' raised to the power of exponent is result); < pre> <p> <strong>Output:</strong> </p> <pre> 2 raised to the power of 3 is 8 </pre> <h3>2. Using Java while Loop</h3> <p>A while loop may similarly be used to achieve the same result by multiplying the base many times.</p> <p> <strong>PowerOfNumber2.java</strong> </p> <pre> public class PowerOfNumber2 { public static void main(String[] args) { int base = 2; int exponent = 3; int result = 1; int power=3; while (exponent &gt; 0) { result *= base; exponent--; } System.out.println(base + &apos; raised to the power of &apos; + power + &apos; is &apos; + result); } } </pre> <p> <strong>Output:</strong> </p> <pre> 2 raised to the power of 3 is 8 </pre> <h3>3. Using Recursion:</h3> <p>Recursion is the process of breaking down an issue into smaller sub-problems. Here&apos;s an example of how recursion may be used to compute a number&apos;s power.</p> <p> <strong>PowerOfNumber3.java</strong> </p> <pre> public class PowerOfNumber3 { public static void main(String[] args) { int base = 2; int exponent = 3; int result = power(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is &apos; + result); } public static int power(int base, int exponent) { if (exponent == 0) { return 1; } else { return base * power(base, exponent - 1); } } } </pre> <p> <strong>Output:</strong> </p> <pre> 2 raised to the power of 3 is 8 </pre> <h3>4. Using Math.pow() Method</h3> <p>The java.lang package&apos;s Math.pow() function computes the power of an integer directly.</p> <p> <strong>PowerOfNumber4.java</strong> </p> <pre> public class PowerOfNumber4 { public static void main(String[] args) { double base = 2.0; double exponent = 3.0; double result = Math.pow(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is &apos; + result); } } </pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 3.0 is 8.0 </pre> <h3>Handling Negative Exponents:</h3> <p>When dealing with negative exponents, the idea of reciprocal powers might be useful. For instance, x^(-n) equals 1/x^n. Here&apos;s an example of dealing with negative exponents.</p> <p> <strong>PowerOfNumber5.java</strong> </p> <pre> public class PowerOfNumber5 { public static void main(String[] args) { double base = 2.0; int exponent = -3; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { if (exponent &gt;= 0) { return calculatePositivePower(base, exponent); } else { return 1.0 / calculatePositivePower(base, -exponent); } } static double calculatePositivePower(double base, int exponent) { double result = 1.0; for (int i = 0; i <exponent; i++) { result *="base;" } return result; < pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of -3 is: 0.125 </pre> <h3>Optimizing for Integer Exponents:</h3> <p>When dealing with integer exponents, you may optimize the calculation by iterating only as many times as the exponent value. It decreases the number of unneeded multiplications.</p> <p> <strong>PowerOfNumber6.java</strong> </p> <pre> public class PowerOfNumber6 { public static void main(String[] args) { double base = 2.0; int exponent = 4; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; for (int i = 0; i <exponent; i++) { result *="base;" } return result; < pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 4 is: 16.0 </pre> <h3>5. Using Bit Manipulation to Calculate Binary Exponents:</h3> <p>Bit manipulation can be used to better improve integer exponents. To do fewer multiplications, an exponent&apos;s binary representation might be used.</p> <p> <strong>PowerOfNumber7.java</strong> </p> <pre> public class PowerOfNumber7 { public static void main(String[] args) { double base = 2.0; int exponent = 5; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; while (exponent &gt; 0) { if ((exponent &amp; 1) == 1) { result *= base; } base *= base; exponent &gt;&gt;= 1; } return result; } } </pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 5 is: 32.0 </pre> <hr></exponent;></pre></exponent;></pre></exponent;>

2. Javan käyttö silmukan aikana

Samalla tavalla voidaan käyttää while-silmukkaa saman tuloksen saavuttamiseksi kertomalla kanta monta kertaa.

PowerOfNumber2.java

 public class PowerOfNumber2 { public static void main(String[] args) { int base = 2; int exponent = 3; int result = 1; int power=3; while (exponent &gt; 0) { result *= base; exponent--; } System.out.println(base + &apos; raised to the power of &apos; + power + &apos; is &apos; + result); } } 

Lähtö:

math.random java
 2 raised to the power of 3 is 8 

3. Rekursion käyttö:

Rekursio on prosessi, jossa ongelma jaetaan pienempiin osaongelmiin. Tässä on esimerkki siitä, kuinka rekursiota voidaan käyttää luvun tehon laskemiseen.

PowerOfNumber3.java

 public class PowerOfNumber3 { public static void main(String[] args) { int base = 2; int exponent = 3; int result = power(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is &apos; + result); } public static int power(int base, int exponent) { if (exponent == 0) { return 1; } else { return base * power(base, exponent - 1); } } } 

Lähtö:

 2 raised to the power of 3 is 8 

4. Math.pow()-menetelmän käyttäminen

Java.lang-paketin Math.pow()-funktio laskee kokonaisluvun potenssin suoraan.

PowerOfNumber4.java

piilotetut sovellukset tällä laitteella
 public class PowerOfNumber4 { public static void main(String[] args) { double base = 2.0; double exponent = 3.0; double result = Math.pow(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is &apos; + result); } } 

Lähtö:

 2.0 raised to the power of 3.0 is 8.0 

Negatiivisten eksponentien käsittely:

Kun käsitellään negatiivisia eksponenteja, ajatus vastavuoroisista voimista voi olla hyödyllinen. Esimerkiksi x^(-n) on yhtä kuin 1/x^n. Tässä on esimerkki negatiivisten eksponentien käsittelystä.

PowerOfNumber5.java

 public class PowerOfNumber5 { public static void main(String[] args) { double base = 2.0; int exponent = -3; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { if (exponent &gt;= 0) { return calculatePositivePower(base, exponent); } else { return 1.0 / calculatePositivePower(base, -exponent); } } static double calculatePositivePower(double base, int exponent) { double result = 1.0; for (int i = 0; i <exponent; i++) { result *="base;" } return result; < pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of -3 is: 0.125 </pre> <h3>Optimizing for Integer Exponents:</h3> <p>When dealing with integer exponents, you may optimize the calculation by iterating only as many times as the exponent value. It decreases the number of unneeded multiplications.</p> <p> <strong>PowerOfNumber6.java</strong> </p> <pre> public class PowerOfNumber6 { public static void main(String[] args) { double base = 2.0; int exponent = 4; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; for (int i = 0; i <exponent; i++) { result *="base;" } return result; < pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 4 is: 16.0 </pre> <h3>5. Using Bit Manipulation to Calculate Binary Exponents:</h3> <p>Bit manipulation can be used to better improve integer exponents. To do fewer multiplications, an exponent&apos;s binary representation might be used.</p> <p> <strong>PowerOfNumber7.java</strong> </p> <pre> public class PowerOfNumber7 { public static void main(String[] args) { double base = 2.0; int exponent = 5; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; while (exponent &gt; 0) { if ((exponent &amp; 1) == 1) { result *= base; } base *= base; exponent &gt;&gt;= 1; } return result; } } </pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 5 is: 32.0 </pre> <hr></exponent;></pre></exponent;>

Optimointi kokonaislukueksponenteille:

Kun käsittelet kokonaislukueksponentteja, voit optimoida laskennan iteroimalla vain eksponentin arvon verran. Se vähentää tarpeettomien kertolaskujen määrää.

PowerOfNumber6.java

 public class PowerOfNumber6 { public static void main(String[] args) { double base = 2.0; int exponent = 4; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; for (int i = 0; i <exponent; i++) { result *="base;" } return result; < pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 4 is: 16.0 </pre> <h3>5. Using Bit Manipulation to Calculate Binary Exponents:</h3> <p>Bit manipulation can be used to better improve integer exponents. To do fewer multiplications, an exponent&apos;s binary representation might be used.</p> <p> <strong>PowerOfNumber7.java</strong> </p> <pre> public class PowerOfNumber7 { public static void main(String[] args) { double base = 2.0; int exponent = 5; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; while (exponent &gt; 0) { if ((exponent &amp; 1) == 1) { result *= base; } base *= base; exponent &gt;&gt;= 1; } return result; } } </pre> <p> <strong>Output:</strong> </p> <pre> 2.0 raised to the power of 5 is: 32.0 </pre> <hr></exponent;>

5. Bittimanipuloinnin käyttäminen binäärieksponenttien laskemiseen:

Bittimanipulaatiolla voidaan parantaa kokonaislukueksponentteja. Kertolaskujen vähentämiseksi voidaan käyttää eksponentin binaariesitystä.

null tarkistus javassa

PowerOfNumber7.java

 public class PowerOfNumber7 { public static void main(String[] args) { double base = 2.0; int exponent = 5; double result = calculatePower(base, exponent); System.out.println(base + &apos; raised to the power of &apos; + exponent + &apos; is: &apos; + result); } static double calculatePower(double base, int exponent) { double result = 1.0; while (exponent &gt; 0) { if ((exponent &amp; 1) == 1) { result *= base; } base *= base; exponent &gt;&gt;= 1; } return result; } } 

Lähtö:

 2.0 raised to the power of 5 is: 32.0