How to resolve the algorithm Iterated digits squaring step by step in the Rust programming language

Published on 12 May 2024 09:40 PM
#Rust

How to resolve the algorithm Iterated digits squaring step by step in the Rust programming language

Table of Contents

Problem Statement

If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89: An example in Python:

Or, for much less credit - (showing that your algorithm and/or language is slow): This problem derives from the Project Euler problem 92. For a quick algorithm for this task see the talk page

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Iterated digits squaring step by step in the Rust programming language

Source code in the rust programming language

fn digit_square_sum(mut num: usize) -> usize {
    let mut sum = 0;
    while num != 0 {
        sum += (num % 10).pow(2);
        num /= 10;
    }
    sum
}

fn last_in_chain(num: usize) -> usize {
    match num {
        1 | 89 => num,
        _ => last_in_chain(digit_square_sum(num)),
    }
}

fn main() {
    let count = (1..100_000_000).filter(|&n| last_in_chain(n) == 89).count();
    println!("{}", count);
}


fn dig_sq_sum(mut num : usize ) -> usize {
    let mut sum = 0;
    while num != 0 {
        sum += (num % 10).pow(2);
        num /= 10;
    }
    sum
}

fn last_in_chain(num: usize) -> usize {
    match num {
        0 => 0,
        1 | 89 => num,
        _ => last_in_chain(dig_sq_sum(num)),
    }
}

fn main() {
    let prec: Vec<_> = (0..649).map(|n| last_in_chain(n)).collect();
    let count = (1..100_000_000).filter(|&n| prec[dig_sq_sum(n)] == 89).count();
    println!("{}", count);
}

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