logo

Java の数値のべき乗

このセクションでは、数値のべき乗を決定する Java プログラムを作成します。数値のべき乗を求めるには、数値に指数を掛けます。

例:

底が 5 で指数が 4 であると仮定します。数値の累乗を求めるには、その数値を 4 回掛けます (5 * 5 * 5 * 5 = 625)。

数値のべき乗を決定するにはどうすればよいですか?

  • 基数と指数は読み取るか初期化する必要があります。
  • 別の変数べき乗を取得し、それを 1 に設定して結果を保存します。
  • for または while ループを使用して、底にべき乗を乗算し、結果をべき乗に格納します。
  • 指数がゼロになるまでステップ 3 を繰り返します。
  • 出力を印刷します。

数値のべき乗を求める方法

数値のべき乗を決定するにはいくつかの方法があります。

jsの置換
  1. Java for ループの使用
  2. Java while ループの使用
  3. 再帰の使用
  4. Math.pow() メソッドの使用
  5. ビット操作の使用

1. Java for ループの使用

for ループを使用すると、基数を繰り返し乗算して数値のべき乗を計算できます。

PowerOfNumber1.java

ピーナッツ vs 落花生
 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 while ループの使用

while ループも同様に使用して、基数を何度も乗算することで同じ結果を得ることができます。

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); } } 

出力:

 2 raised to the power of 3 is 8 

3. 再帰の使用:

再帰は、問題をより小さなサブ問題に分解するプロセスです。以下は、再帰を使用して数値のべき乗を計算する方法の例です。

PowerOfNumber3.java

ランタイムエラー
 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); } } } 

出力:

 2 raised to the power of 3 is 8 

4. Math.pow() メソッドの使用

java.lang パッケージの Math.pow() 関数は、整数のべき乗を直接計算します。

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

出力:

 2.0 raised to the power of 3.0 is 8.0 

負の指数の処理:

負の指数を扱う場合、逆べき乗の考え方が役立つ場合があります。たとえば、x^(-n) は 1/x^n と等しくなります。以下は負の指数を扱う例です。

PowerOfNumber5.java

C++で優先キュー
 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;>

整数の指数の最適化:

整数の指数を扱う場合、指数の値と同じ回数だけ反復することで計算を最適化できます。不必要な乗算の数が減ります。

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. ビット操作を使用して 2 進指数を計算する:

ビット操作を使用すると、整数の指数を改善することができます。実行する乗算の​​数を減らすために、指数のバイナリ表現が使用される場合があります。

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; } } 

出力:

 2.0 raised to the power of 5 is: 32.0