• 欧拉计划 第5题


    2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

    What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

    internal class Program
        {
            private static void Main(string[] args)
            {
                /*
                * 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
                * What is the smallest number that is evenly divisible by all of the numbers from 1 to 20? 
                 */
    
                int[] numbers = Enumerable.Range(1, 20).ToArray();
                int prime = 2;
                int nextPrime = 2;
                int nextPrimePow2 = 2*2;
                for (int i = 1; nextPrimePow2 < 20; i++)
                {
                    int n = numbers[i];
                    if (n == 0) continue;
                    for (int j = i + 1; j < numbers.Length; j++)
                    {
                        int k = numbers[j];
                        if (k%prime == 0)
                        {
                            numbers[j] = 0;
                            continue;
                        }
                        if (prime == nextPrime) nextPrime = k;
                    }
                    prime = nextPrime;
                    nextPrimePow2 = nextPrime*nextPrime;
                }
                var primes = numbers.Where(p => p != 0 && p != 1).ToArray();
                var primeNums = new int[primes.Length];
    
                numbers = Enumerable.Range(1, 20).ToArray();
    
    
                for (int i = 0; i < numbers.Length; i++)
                {
                    int n = numbers[i];
                    for (int j = 0; j < primes.Length; j++)
                    {
                        int item = primes[j];
                        int k = 0;
                        for (; ; k++, n /= item)
                        {
                            if (n % item != 0) break;
                        }
                        if (primeNums[j] < k) primeNums[j] = k;
                    }
                }
                long result = 1;
                for (int i = 0; i < primes.Length; i++)
                {
                    Console.WriteLine("{0}*{1}", primes[i], primeNums[i]);
                    result *= (long)Math.Pow(primes[i],primeNums[i]);
                }
                Console.WriteLine(result);
            }
        }
    

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  • 原文地址:https://www.cnblogs.com/kiminozo/p/2156292.html
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