# C++ solution to Project Euler Problem 10

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

Running time: ~5 secs

Assessment: This solution is in C++ for some reason. Oh well. Anyway, it works fine, but like most of my other early factorization solutions, it doesn’t memoize anything, which means lots of redundant work, which makes for very long execution times. It’s pretty bad all around.

```#include <iostream>
#include <cmath>

using namespace std;

long long is_prime(long long n)
{
//returns 0 if not prime, 1 if prime
if (n % 2 == 0)
return 0;
for (long long i = 3; i <= (pow(n,0.5)); i += 2)
{
if (n % i == 0)
return 0;
}
return 1;
}

int main()
{
unsigned long long sum = 0;

for (int i = 3; i < 2000000; i += 2)
{
if (is_prime(i))
{
sum += i;
}
}

cout << sum + 2;

return 0;
}```

# Java solution to Project Euler Problem 8

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Running time: 35ms

Assessment: Pretty straightforward; just iterate through the number character by character and check it.

```public class Problem008
{
public static int checkInt(String s)
{
int product = 1;

for (int i = 0; i < 5; i++)
{
Character c = new Character(s.charAt(i));
String tmp = c.toString();
int temp = Integer.parseInt(tmp);
product *= temp;
}
return product;
}

public static void main(String[] args)
{
long begin = System.currentTimeMillis();
String BigNum = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
String snip;
int largest = 0;

for (int i = 0; i <= (BigNum.length()-5); i++)
{
snip = "";

for (int j = 0; j < 5; j++)
{
char c = BigNum.charAt(i+j);
snip += c;
}

if (checkInt(snip) > largest)
largest = checkInt(snip);
}
long end = System.currentTimeMillis();
System.out.println(largest);
System.out.println(end-begin + "ms");
}
}```

# Java solution to Project Euler Problem 7

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?

Running time: 10010ms (10 sec)

Assessment: This is a classic example of a bad factorization algorithm, because the results aren’t memoized in any way. I promise my solutions get a lot better right around Problem 21.

```public class Problem007
{
private static boolean isPrime(long n)
{
//returns 0 if not prime, 1 if prime
if ((n % 2 == 0)&&(n != 2))
return false;		// is even, therefore not prime
for (long i = 3; i <= (n^(1/2)+1); i += 2)	// Skip all the even numbers
{
if (n % i == 0)
return false;	//not prime
}
return true;
}

public static void main(String[] args)
{
long begin = System.currentTimeMillis();
int NumPrimes = 1;
long i = 2;
while(NumPrimes <= 10001)
{
if (isPrime(i))
{
i++;
NumPrimes++;
}
else i++;
}
long end = System.currentTimeMillis();
System.out.println(i);
System.out.println(end-begin + "ms");
}
}```

# Java solution to Project Euler Problem 6

The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + … + 10^2 = 385

The square of the sum of the first ten natural numbers is,
(1 + 2 + … + 10)^2 = 55^2 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is
3025 – 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Running time: 0ms

Assessment: I exhibited symptoms of brain inversion the first time I read the problem description. I was very tired when I wrote this. How do I know? There are zombies in my code. ```public class Problem006
{
private static long sum_squares(int n)
{
int total = 0;
for (int i = 1; i <= n; i++)
total += (i*i);
System.out.println("sum_squares() = " + total);
return total;
}

private static long squares_sum(int n)
{
long total = 0;
for (int i = 1; i <= n; i++)
total += i;
total *= total;
System.out.println("squares_sum() = " + total);
return total;
}

public static void main(String args[])
{
long begin = System.currentTimeMillis();
long omg_zombies = squares_sum(100) - sum_squares(100);
long end = System.currentTimeMillis();
System.out.println(omg_zombies);
System.out.println(end-begin + "ms");
}
}```

# Java solution to Project Euler Problem 5

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?

Running time: 125ms

Assessment: LOL.

```public class Problem005
{
public static void main(String[] args)
{
long begin = System.currentTimeMillis();
int i = 20;
while (true)
{
if (	(i % 1 == 0) &&
(i % 2 == 0) &&
(i % 3 == 0) &&
(i % 4 == 0) &&
(i % 5 == 0) &&
(i % 6 == 0) &&
(i % 7 == 0) &&
(i % 8 == 0) &&
(i % 9 == 0) &&
(i % 10 == 0) &&
(i % 11 == 0) &&
(i % 12 == 0) &&
(i % 13 == 0) &&
(i % 14 == 0) &&
(i % 15 == 0) &&
(i % 16 == 0) &&
(i % 17 == 0) &&
(i % 18 == 0) &&
(i % 19 == 0) &&
(i % 20 == 0) )
{
break;
}
i += 20;
}
long end = System.currentTimeMillis();
System.out.println(i);
System.out.println(end-begin + "ms");
}
}```

# Java solution to Project Euler Problem 4

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.

Find the largest palindrome made from the product of two 3-digit numbers.

Running time: 5ms

Assessment: Despite finishing quickly, this is a bad solution, because it returns the first time it reaches a palindrome. The problem is that there’s no guarantee that the return value is, in fact, the correct answer. Using different input criteria could easily break it. It so happens that the returned value of this code is correct, but this was more luck than skill.

```public class Problem004
{
private static int createPalindrome()
{
int total = 0;
for (int i = 999; i >= 900; i--)
{
for (int j = 999; j >= 900; j--)
{
total = i * j;
String s = Integer.toString(total);
String s2 = new StringBuffer(s).reverse().toString();
if (s.equals(s2))
return total;
}
}
return 0;
}

public static void main(String[] args)
{
long begin = System.currentTimeMillis();
int answer = createPalindrome();
long end = System.currentTimeMillis();
System.out.println(answer);
System.out.println(end-begin + "ms");
}
}```

# Java solution to Project Euler Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

Running time: 2ms

Assessment: This was when I switched from C++ to Java, and this solution is a lot better than my C++ solution. This one finishes, and finishes pretty quickly.

```public class Problem003b
{
public static void main(String[] args)
{
long begin = System.currentTimeMillis();
long n=600851475143L;
for (long i = 2; i <= n; i++)
{
if (n % i==0)
{
System.out.println(i);
n = n / i;
i = 2;
}
}
long end = System.currentTimeMillis();
System.out.println(end-begin + "ms");
}
}```

# C++ solution to Project Euler Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

Running time: Unknown, never finishes

Assessment: Brute forced, naive, ugly, doesn’t finish. It was really hard to post this as-is without self-editing to make me look less like a nub, but the point of posting these is to show the evolution of thought processes and problem-solving over time.

The answer is displayed more or less right away, but the code never exits, so you’re not sure if the last answer displayed before you get annoyed and hit Ctrl-C is the correct one. (It is.) I never had the patience to let it finish, let alone measure the runtime. I would approach the problem completely differently–as you’ll see in some of the later factorization examples–if I were to re-write it today.

1. `#include <iostream>`
2. `using namespace std;`
3. ` `
4. `long long is_prime(long long n)`
5. `{`
6. `	//returns 0 if not prime, 1 if prime`
7. `	if (n % 2 == 0)`
8. `		return 0;		// is even, therefore not prime`
9. `	for (long long i = 3; i <= ((n / 2) + 1); i += 2)	// Skip all the even numbers`
10. `	{`
11. `		if (n % i == 0)`
12. `			return 0;	//not prime`
13. `	}`
14. `	return 1;`
15. `}`
16. ` `
17. `long long find_bigprime(long long n)`
18. `{`
19. `	long long factor = 0;`
20. `	for(long long i = 3; i <= n; i += 2)`
21. `	{`
22. `		if (n % i == 0)`
23. `		{`
24. `			if (is_prime(i))`
25. `			{`
26. `				factor = i;`
27. `				cout << factor << endl;`
28. `			}`
29. `		}`
30. `	}`
31. `	return factor;`
32. `}`
33. ` `
34. `int main()`
35. `{`
36. `	long long input = 600851475143;`
37. `	cout << find_bigprime(input);`
38. ` `
39. `	return 0;`
40. `}`

# C++ solution to Project Euler Problem 2

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Running time: Unknown

Assessment: Inelegant.

This took me an embarrassingly long time to get right. With the exception of Problem 1, I hadn’t written any code in a few years. As noted before, I hadn’t started measuring execution time yet, so I’m not sure how long it took to run, but it’s basically instantaneous.

1. `#include <iostream>`
2. `using namespace std;`
3. ` `
4. `int main()`
5. `{`
6. `	int total = 0;`
7. ` `
8. `	//3 variables to create the Fibonacci sequence from 1 to X`
9. `	int one = 1;`
10. `	int two = 2;`
11. `	int three = 0;`
12. ` `
13. `	while (two <= 4000000)`
14. `	{`
15. `		three = one + two;`
16. `		if (two % 2 == 0)`
17. `			total += two;`
18. `		one = two;`
19. `		two = three;`
20. `	}`
21. ` `
22. `	cout << total;`
23. ` `
24. `	return 0;`
25. `}`

# C++ solution to Project Euler Problem 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Running time: Unknown

Assessment: First code I’d written in 7-8 years. I hadn’t started measuring execution time yet, so I’m not sure how long it took to run, but it’s basically instantaneous.

1. `#include <iostream>`
2. `using namespace std;`
3. ` `
4. `int main()`
5. `{`
6. `	int total = 0;`
7. `	for (int i = 0; i < 1000; i++)`
8. `	{`
9. `		if( (i % 3 == 0) || (i % 5 == 0) )`
10. `			total += i;`
11. `	}`
12. `	cout << total;	`
13. `	return 0;`
14. `}`