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Šiame skyriuje parašysime Java programas skaičiaus galiai nustatyti. Norėdami gauti skaičiaus laipsnį, skaičių padauginkite iš jo eksponento.

Pavyzdys:

Tarkime, kad bazė yra 5, o rodiklis yra 4. Norėdami gauti skaičiaus laipsnį, padauginkite jį iš savęs keturis kartus, t.y. (5 * 5 * 5 * 5 = 625).

Kaip nustatyti skaičiaus galią?

• Bazė ir eksponentas turi būti nuskaityti arba inicijuoti.
• Paimkite kitą kintamąjį laipsnį ir nustatykite jį į 1, kad išsaugotumėte rezultatą.
• Padauginkite bazę iš galios ir išsaugokite rezultatą galioje naudodami for arba while kilpą.
• Kartokite 3 veiksmą, kol eksponentas bus lygus nuliui.
• Spausdinkite išvestį.

Skaičiaus galios nustatymo metodai

Yra keletas būdų, kaip nustatyti skaičiaus galią:

paveldėjimo java

1. „Java for Loop“ naudojimas
2. „Java“ naudojimas „loop“ metu
3. Rekursijos naudojimas
4. Naudojant Math.pow() metodą
5. Naudojant bitų manipuliavimą

1. Java for Loop naudojimas

Ciklas for gali būti naudojamas skaičiaus galiai apskaičiuoti pakartotinai dauginant bazę iš savęs.

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. Java naudojimas, kai ciklas

Tam pačiam rezultatui pasiekti galima panašiai naudoti o kilpą, daug kartų padauginus bazę.

PowerOfNumber2.java

privati ​​vs vieša java

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

Išvestis:

 2 raised to the power of 3 is 8 

3. Rekursijos naudojimas:

Rekursija yra problemos suskaidymo į smulkesnes problemas procesas. Štai pavyzdys, kaip rekursija gali būti naudojama skaičiaus galiai apskaičiuoti.

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 + ' raised to the power of ' + exponent + ' is ' + result); } public static int power(int base, int exponent) { if (exponent == 0) { return 1; } else { return base * power(base, exponent - 1); } } } 

Išvestis:

 2 raised to the power of 3 is 8 

4. Naudojant Math.pow() metodą

Paketo java.lang funkcija Math.pow() tiesiogiai apskaičiuoja sveikojo skaičiaus galią.

tinklų tipai

PowerOfNumber4.java

 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 + ' raised to the power of ' + exponent + ' is ' + result); } } 

Išvestis:

 2.0 raised to the power of 3.0 is 8.0 

Neigiamų eksponentų tvarkymas:

Kalbant apie neigiamus eksponentus, gali būti naudinga abipusių galių idėja. Pavyzdžiui, x^(-n) lygus 1/x^n. Štai pavyzdys, kaip elgtis su neigiamais eksponentais.

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;>

Optimizavimas sveikiesiems rodikliams:

Kai dirbate su sveikaisiais rodikliais, galite optimizuoti skaičiavimą kartodami tik tiek kartų, kiek yra eksponento reikšmė. Tai sumažina nereikalingų padauginimų skaičių.

PowerOfNumber6.java

kokie mėnesiai yra Q3

 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. Bitų manipuliavimo naudojimas dvejetainiams eksponentams apskaičiuoti:

Bitų manipuliavimas gali būti naudojamas norint geriau pagerinti sveikųjų skaičių eksponentus. Norint atlikti mažiau daugybos, gali būti naudojamas eksponento dvejetainis vaizdas.

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 + ' raised to the power of ' + exponent + ' is: ' + result); } static double calculatePower(double base, int exponent) { double result = 1.0; while (exponent > 0) { if ((exponent & 1) == 1) { result *= base; } base *= base; exponent >>= 1; } return result; } } 

Išvestis:

 2.0 raised to the power of 5 is: 32.0