Problem Description<\/h2>\n<\/header>\n\nThe Euler’s Totient function (\u03c6(n)) represents the number of positive integers that are coprime to n for a given natural number n. For example, \u03c6(1) = 1, \u03c6(2) = 1, \u03c6(3) = 2, \u03c6(4) = 2, \u03c6(5) = 4. The goal of this problem is to write a program that calculates \u03c6(n) for a given natural number n.<\/p>\n<\/section>\n\nProblem: Calculate \u03c6(n)<\/h2>\n
When a natural number n is given, write a Java program to calculate \u03c6(n) for n. Note that n is an integer between 1 and 10^6 inclusive.<\/p>\n
Input Format<\/h3>\n\n- Natural number n is provided in the first line.<\/li>\n<\/ul>\n
Output Format<\/h3>\n\n- Output the value of \u03c6(n) for n.<\/li>\n<\/ul>\n<\/section>\n
\nProblem Solving Process<\/h2>\n
To solve the problem, one must understand the definition and properties of Euler’s Totient function accurately and design an algorithm to calculate it efficiently.<\/p>\n
Definition of Euler’s Totient Function<\/h3>\n
The Euler’s Totient function is calculated as follows:<\/p>\n
\n\u03c6(n) = n * (1 - 1\/p1) * (1 - 1\/p2) * ... * (1 - 1\/pk)\n <\/pre>\nHere, p1, p2, …, pk are the distinct prime factors of n. That is, the result can be derived from the number of primes that appear in the prime factorization of n and their product.<\/p>\n
Calculating \u03c6(n) through Prime Factorization<\/h3>\n
The steps to use this equation to calculate \u03c6(n) in Java are as follows:<\/p>\n
\n- Set the initial value of n received as input.<\/li>\n
- Iterate from 2 to the square root of n to find primes.<\/li>\n
- For each prime p, check if n is divisible by p, and if it is, divide n by p and multiply \u03c6(n) by (1 – 1\/p).<\/li>\n
- Finally, repeat until n becomes 1.<\/li>\n
- Output the calculated value of \u03c6(n).<\/li>\n<\/ol>\n<\/section>\n
\nJava Code Example<\/h2>\n
Below is an example of Java code that calculates \u03c6(n):<\/p>\n
\nimport java.util.Scanner;\n\npublic class EulerTotient {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int n = scanner.nextInt();\n int originalN = n;\n double result = n;\n\n for (int i = 2; i * i <= n; i++) {\n if (n % i == 0) {\n while (n % i == 0) {\n n \/= i;\n }\n result *= (1.0 - (1.0 \/ (double)i));\n }\n }\n\n if (n > 1) {\n result *= (1.0 - (1.0 \/ (double)n));\n }\n\n System.out.println((int)result);\n }\n}\n <\/pre>\nThis code calculates and outputs \u03c6(n) for the value of n entered by the user. It has calculated \u03c6(n) in cases where it can be divided by primes and the response is provided accordingly.<\/p>\n<\/section>\n\nEfficiency Analysis<\/h2>\n
The time complexity of this algorithm is O(\u221an) because it needs to iterate up to the square root of n to find primes. This method can handle cases where n is as large as 10^6 without issues. Moreover, the space complexity is O(1), minimizing additional memory usage, making it efficient.<\/p>\n<\/section>\n\nConclusion<\/h2>\n
The Euler’s Totient function is one of the very important concepts in number theory and algorithm problems. Through the process of solving this problem, one can enhance their understanding of primes, prime factorization, and basic mathematical principles. Remember that this is a useful problem for preparing for coding tests using Java.<\/p>\n<\/section>\n
The Euler’s Totient function (\u03c6(n)) represents the number of positive integers that are coprime to n for a given natural number n. For example, \u03c6(1) = 1, \u03c6(2) = 1, \u03c6(3) = 2, \u03c6(4) = 2, \u03c6(5) = 4. The goal of this problem is to write a program that calculates \u03c6(n) for a given natural number n.<\/p>\n<\/section>\n When a natural number n is given, write a Java program to calculate \u03c6(n) for n. Note that n is an integer between 1 and 10^6 inclusive.<\/p>\n To solve the problem, one must understand the definition and properties of Euler’s Totient function accurately and design an algorithm to calculate it efficiently.<\/p>\n The Euler’s Totient function is calculated as follows:<\/p>\n Here, p1, p2, …, pk are the distinct prime factors of n. That is, the result can be derived from the number of primes that appear in the prime factorization of n and their product.<\/p>\n The steps to use this equation to calculate \u03c6(n) in Java are as follows:<\/p>\n Below is an example of Java code that calculates \u03c6(n):<\/p>\n This code calculates and outputs \u03c6(n) for the value of n entered by the user. It has calculated \u03c6(n) in cases where it can be divided by primes and the response is provided accordingly.<\/p>\n<\/section>\n The time complexity of this algorithm is O(\u221an) because it needs to iterate up to the square root of n to find primes. This method can handle cases where n is as large as 10^6 without issues. Moreover, the space complexity is O(1), minimizing additional memory usage, making it efficient.<\/p>\n<\/section>\n The Euler’s Totient function is one of the very important concepts in number theory and algorithm problems. Through the process of solving this problem, one can enhance their understanding of primes, prime factorization, and basic mathematical principles. Remember that this is a useful problem for preparing for coding tests using Java.<\/p>\n<\/section>\nProblem: Calculate \u03c6(n)<\/h2>\n
Input Format<\/h3>\n
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Output Format<\/h3>\n
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Problem Solving Process<\/h2>\n
Definition of Euler’s Totient Function<\/h3>\n
\n\u03c6(n) = n * (1 - 1\/p1) * (1 - 1\/p2) * ... * (1 - 1\/pk)\n <\/pre>\n
Calculating \u03c6(n) through Prime Factorization<\/h3>\n
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Java Code Example<\/h2>\n
\nimport java.util.Scanner;\n\npublic class EulerTotient {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int n = scanner.nextInt();\n int originalN = n;\n double result = n;\n\n for (int i = 2; i * i <= n; i++) {\n if (n % i == 0) {\n while (n % i == 0) {\n n \/= i;\n }\n result *= (1.0 - (1.0 \/ (double)i));\n }\n }\n\n if (n > 1) {\n result *= (1.0 - (1.0 \/ (double)n));\n }\n\n System.out.println((int)result);\n }\n}\n <\/pre>\nEfficiency Analysis<\/h2>\n
Conclusion<\/h2>\n