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Snaga broja u Javi

U ovom odjeljku ćemo napisati Java programe za određivanje snage broja. Da biste dobili potenciju broja, pomnožite broj s eksponentom.

Primjer:

Pretpostavimo da je baza 5, a eksponent 4. Da biste dobili potenciju broja, pomnožite ga samim sobom četiri puta, tj. (5 * 5 * 5 * 5 = 625).

Kako odrediti snagu broja?

  • Baza i eksponent trebaju se očitati ili inicijalizirati.
  • Uzmite drugu varijablu snage i postavite je na 1 da biste spremili rezultat.
  • Pomnožite bazu s potencijom i pohranite rezultat u potenciju pomoću petlje for ili while.
  • Ponovite korak 3 dok eksponent ne bude jednak nuli.
  • Ispišite izlaz.

Metode određivanja snage broja

Postoji nekoliko metoda za određivanje snage broja:

ponavljanje karte u Javi
  1. Korištenje Jave za petlju
  2. Korištenje Java while petlje
  3. Korištenje rekurzije
  4. Korištenje metode Math.pow().
  5. Korištenje Bit Manipulacije

1. Korištenje Jave za petlju

For petlja se može koristiti za izračunavanje snage broja uzastopnim množenjem baze samom sobom.

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. Korištenje Java while petlje

Slično se može koristiti while petlja za postizanje istog rezultata množenjem baze mnogo puta.

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); } } 

Izlaz:

pretvarač niza u int
 2 raised to the power of 3 is 8 

3. Korištenje rekurzije:

Rekurzija je proces rastavljanja problema na manje podprobleme. Evo primjera kako se rekurzija može koristiti za izračunavanje potencije broja.

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); } } } 

Izlaz:

 2 raised to the power of 3 is 8 

4. Korištenje metode Math.pow().

Funkcija Math.pow() paketa java.lang izravno izračunava snagu cijelog broja.

PowerOfNumber4.java

ime grada u SAD-u
 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); } } 

Izlaz:

 2.0 raised to the power of 3.0 is 8.0 

Rukovanje negativnim eksponentima:

Kad se radi o negativnim eksponentima, ideja recipročnih potencija mogla bi biti korisna. Na primjer, x^(-n) jednako je 1/x^n. Evo primjera postupanja s negativnim eksponentima.

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;>

Optimiziranje za cjelobrojne eksponente:

Kada radite s cjelobrojnim eksponentima, možete optimizirati izračun ponavljanjem samo onoliko puta koliko je eksponentna vrijednost. Smanjuje broj nepotrebnih množenja.

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. Korištenje bitne manipulacije za izračun binarnih eksponenata:

Manipulacija bitovima može se koristiti za bolje poboljšanje cjelobrojnih eksponenata. Za manje množenja može se koristiti binarna reprezentacija eksponenta.

usporediti u nizu

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; } } 

Izlaz:

 2.0 raised to the power of 5 is: 32.0