Euler #8, #9 y #10

https://projecteuler.net/problem=8

Sencillo problema, lo más fácil es tratar el número tan largo que proponen como un string e ir tomando "porciones" de 13 caracteres y sumándolas, guardándonos la mayor suma. Implementación en Python.

   1 
   2 number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
   3 
   4 biggest = ""
   5 product = 0
   6 length = 13
   7 
   8 count = 0
   9 portion = number[count:count+length]
  10 while len(portion) == length:
  11     temp = 1
  12     for d in portion:
  13         temp *= int(d)
  14     if temp > product:
  15         product = temp
  16         biggest = portion
  17     count += 1
  18     portion = number[count:count+length]
  19 
  20 print ("Biggest: " + biggest + ", product: " + str(product))
  21 

---

https://projecteuler.net/problem=9

Implementación en JavaScript

   1 var SUM = 1000
   2 for (var a = 1; a < SUM; a++) {
   3   for (var b = a + 1; b < SUM; b++) {
   4     c2 = a*a + b*b;
   5     c = Math.sqrt(c2)
   6     if (Math.round(c) == c) { // Pythagorean triplet
   7       if (a + b +c == SUM) {
   8         console.log("Triplet: ", a, b, c, "Product: ", a*b*c);
   9         process.exit(0);
  10       }
  11     }
  12   }
  13 }
  14 

---

https://projecteuler.net/problem=10

Utilizamos la función sum de Python para efectuar la suma de todos los números. El problema principal es tener una función generadora de números primos. La que se propone es una sencilla criba de Eratóstenes.

   1 import math
   2 
   3 def isPrime(n):
   4   if n == 2:
   5       return True
   6   if n < 2 or n % 2 == 0:
   7       return False;
   8   max = int(math.sqrt(n));
   9   for i in range(3, max+1, 2):
  10       if n % i == 0:
  11           return False
  12   return True
  13 
  14 
  15 def primeGenerator(n):
  16     """Generates prime numbers < n"""
  17     primes = [2]
  18     if n > 2:
  19         candidates = [c for c in range(3, n, 2)]
  20         for test in range(3, int(math.sqrt(n))+1, 2):
  21             for c in candidates:
  22                 if c > test and c % test == 0:
  23                     candidates.remove(c)
  24         primes += candidates
  25     return primes
  26 
  27 if __name__ == '__main__':
  28     primes = primeGenerator(2000000)
  29     print sum(primes)
  30 

 

Euler #5, #6 y #7

https://projecteuler.net/problem=5

   1 #!/usr/bin/env python3
   2 
   3 divisors = range(1, 20);
   4 test = divisors[-1]
   5 control = True
   6 while control:
   7     divisible = True
   8     for d in divisors:
   9         if test % d != 0:
  10             divisible = False
  11             break
  12 
  13     if divisible:
  14         control = False
  15     else:
  16         test += 1
  17 
  18 print (test)
  19 

---

https://projecteuler.net/problem=6

JavaScript:

   1 
   2 var sum = 0;
   3 var squares = 0;
   4 for (var i = 0; i <= 100; i++) {
   5   sum += i;
   6   squares += i*i;
   7 }
   8 console.log(sum * sum - squares);
   9 

---

https://projecteuler.net/problem=7

PHP:

   1 <?php
   2 // https://projecteuler.net/problem=7
   3 
   4 function isPrime($num) {
   5   if ($num == 2) return true;
   6   if ($num < 2 || $num % 2 == 0) return false;
   7   $max = (int) sqrt($num);
   8   for ($i = 3; $i <= $max; $i +=2) {
   9       if ($num % $i == 0) {
  10           return false;
  11       }
  12   }
  13   return true;
  14 }
  15 
  16 $counter = 0;
  17 $num = 2;
  18 $stop = false;
  19 while (true) {
  20   if (isPrime($num)) {
  21     $counter++;
  22     if ($counter == 10001) {
  23       echo "$num\n";
  24       break;
  25     }
  26   }
  27   $num++;
  28 }
  29 

Euler #4

https://projecteuler.net/problem=4

Python:

   1 #!/usr/bin/env python3
   2 
   3 def isPalindromic(arg):
   4     arg = str(arg)
   5     cut = int(len(arg)/2)
   6     (first, last) = (arg[:cut], arg[cut:][::-1])
   7     if (len(last) == len(first) + 1):
   8         last = last[:-1]
   9     return first == last
  10     
  11 biggest = 0    
  12 for x in range(100, 1000):
  13     for y in range(100, 1000):
  14         num = x * y
  15         if isPalindromic(num) and num > biggest:
  16             biggest = num
  17             
  18 print (biggest)

Euler #3

Seguimos resolviendo problemas del proyecto Euler, vamos con el tercero.

https://projecteuler.net/problem=3

En Java.

Implementaremos el método más sencillo, división por tentativa.

   1 
   2 public class EulerSqrtRev_3 {
   3 
   4     public static boolean isPrime(long number) {
   5         if (number == 2) return true;
   6         if (number < 2 || number % 2 == 0) return false;
   7         double max = Math.sqrt(number);
   8         for (long i = 3; i <= max; i +=2) {
   9             if (number % i == 0) {
  10                 return false;
  11             }
  12         }
  13         return true;
  14     }
  15     
  16     public static void main(String args[]) {
  17         final long number =  600851475143L;
  18         long max = number/2;
  19         for (long i = 2; i < max; i++) {
  20             if (number % i == 0 && isPrime(i)) {
  21                 System.out.println(i);
  22             }
  23         }
  24     }
  25 
  26 }

 

Euler #1 y #2

Hay que mantenerse en forma...

https://projecteuler.net/problem=1

Python:

   1 #!/usr/bin/env python3
   2 
   3 sum = 0
   4 for n in range(1, 1000):
   5     if n % 3 == 0 or n % 5 == 0:
   6         sum += n
   7 
   8 print (sum)
   9 

JavaScript:

   1 var sum = 0;
   2 for (var i=1; i<1000; i++) {
   3   if (i % 3 == 0 || i % 5 == 0) {
   4     sum += i;
   5   }
   6 }
   7 console.log(sum)
   8 

---

https://projecteuler.net/problem=2

PHP (iterativo):

   1 <?php
   2 
   3 function fib($max) {
   4   $ret = array(1, 2);
   5   $x = 3;
   6   while (true) {
   7     $new = $ret[$x-2] + $ret[$x-3];
   8     if ($new <= $max) {
   9       $ret []= $new;
  10       $x++;
  11     } else {
  12       break;
  13     }
  14   }
  15   return $ret;
  16 }
  17 
  18 
  19 $sum = 0;
  20 foreach (fib(4000000) as $n) {
  21   if ($n % 2 == 0) {
  22     $sum += $n;
  23   }
  24 }
  25 
  26 echo "$sum\n";
  27