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## Project Euler Problem#11 solution in C++

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08

49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65

52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91

22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80

24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50

32 98 81 28 64 23 67 10 **26** 38 40 67 59 54 70 66 18 38 64 70

67 26 20 68 02 62 12 20 95 **63** 94 39 63 08 40 91 66 49 94 21

24 55 58 05 66 73 99 26 97 17 **78** 78 96 83 14 88 34 89 63 72

21 36 23 09 75 00 76 44 20 45 35 **14** 00 61 33 97 34 31 33 95

78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92

16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57

86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58

19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40

04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66

88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69

04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36

20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16

20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54

01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 63 78 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 2020 grid?

Solution:

#include<iostream> const int num = 20; void multiply(long long&, int [][num], int [], int, int, int ,int); int main() { int array[num][num] = { 8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8, 49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0, 81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65, 52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91, 22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80, 24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50, 32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70, 67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21, 24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72, 21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95, 78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92, 16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57, 86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58, 19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40, 4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66, 88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69, 4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36, 20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16, 20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54, 1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48 }; long long maxNum = 1; int bArray[4]; for (int i = 0; i < num; i++) { for (int j = 0; j < num; j++) { multiply(maxNum, array, bArray, i, j, 1, 0); multiply(maxNum, array, bArray, i, j, 0, 1); multiply(maxNum, array, bArray, i, j, 1, 1); multiply(maxNum, array, bArray, i, j, -1, 1); } } std::cout << maxNum << " = " << bArray[0] << " * " << bArray[1] << " * " << bArray[2] << " * " << bArray[3] << std::endl; return 0; } void multiply(long long& maxNum, int array[][num], int bArray[], int x, int y, int dx, int dy) { long long tempNum = 1; for (int i = 0; i < 4 && x < num && x >=0 && y < num && y >= 0; i++, x += dx, y += dy) tempNum *= array[x][y]; if (maxNum < tempNum) { maxNum = tempNum; bArray[0] = array[x - dx][y - dy]; bArray[1] = array[x - dx * 2][y - dy * 2]; bArray[2] = array[x - dx * 3][y - dy * 3]; bArray[3] = array[x - dx * 4][y - dy * 4]; } }

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## Project Euler Problem#10 solution in C++

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

Find the sum of all the primes below two million.

**Explanation:**

In mathematics, the **sieve of Eratosthenes** (Greek: κόσκινον Ἐρατοσθένους), one of a number of prime number sieves, is a simple, ancient algorithm for finding all prime numbers up to any given limit. It does so by iteratively marking as composite (i.e. not prime) the multiples of each prime, starting with the multiples of 2.^{[1]}

The multiples of a given prime are generated starting from that prime, as a sequence of numbers with the same difference, equal to that prime, between consecutive numbers.^{[1]} This is the sieve’s key distinction from using trial division to sequentially test each candidate number for divisibility by each prime.^{[2]}

The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so).^{[3]} It is named after Eratosthenes of Cyrene, a Greek mathematician; although none of his works have survived, the sieve was described and attributed to Eratosthenes in the *Introduction to Arithmetic* by Nicomachus.^{[4]}

(above explanation taken from Wikipedia)

C++ solution:

*//solution is written in a flexible manner that you could compute sum of n prime numbers.*

#include <iostream>

using namespace std;

int find_prime (long *numArray, int maxNum)

{

long factor = 2; // we will make 2 as the starting point.

numArray[1] = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.

//maxNum is the number till there we can find the prime numbers.

while ((factor * factor) < maxNum)

{

for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.

{

if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we havne’t yet checked it.

{

// we are putting zeros on all multiples of prime numbers, one by one.

if (numArray[i] != factor && (numArray[i]% factor) == 0)

{

numArray[i] = 0;

}

}

}

++factor;

}// after the loop, array will have zeros on all non prime locations and prime numbers.

}

int main ()

{

int prime_count = 0;

int64_t sum = 0;

int maxNum = 0;

int i = 0;

cout << “enter max number for array: “; // you need to test some upper limit values.

cin >> maxNum;

long *myArray = new long [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.

for (int i = 0; i < maxNum; i++)

{

myArray[i]= i;

}

// we will get prime numbers till the maxNum by calling below funtions.

find_prime(myArray, maxNum);

for (int j = 0; j< maxNum; j++)

{

sum += myArray[j];

}

cout << “Sum of first ” << maxNum << ” prime numbers : ” << sum << endl;

}

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## Project Euler Problem#7 solution in C++ (Brute Force)

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 10 001st prime number?

Brute Force Solution:

#include <iostream>

using namespace std;

int find_prime (long *numArray, int maxNum)

{

long factor = 2; // we will make 2 as the starting point.

numArray[1] = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.

//maxNum is the number till there we can find the prime numbers.

while ((factor * factor) < maxNum)

{

for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.

{

if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we havne’t yet checked it.

{

// we are putting zeros on all multiples of prime numbers, one by one.

if (numArray[i] != factor && (numArray[i]% factor) == 0)

{

numArray[i] = 0;

}

}

}

++factor;

}// after the loop, array will have zeros on all non prime locations and prime numbers.

}

int main ()

{

int prime_count = 0;

int nth_prime = 0;

int maxNum = 0;

int i = 0;

cout << “enter max number for array: “; // you need to test some upper limit values.

cin >> maxNum;

cout << endl << “enter nth prime number you are searching for: “;

cin >> nth_prime;

long *myArray = new long [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.

for (int i = 0; i < maxNum; i++)

{

myArray[i]= i;

}

// we will get prime numbers till the maxNum by calling below funtions.

find_prime(myArray, maxNum);

for (; i < maxNum; i++)

{

if (myArray[i] != 0)

{

prime_count++;

}

if (prime_count == nth_prime){break;}

}

cout << “the nth prime number is: ” << “value of prime count is: “<< prime_count << ” – ” << myArray[i] << endl;

}

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## Project Euler Problem# 5 solution in C++

/**

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?

Below is an efficient solution of above problem using C++. Time complexity in the random access machine model is O(n log log n)

operations, a direct consequence of the fact that the prime harmonic series asymptotically approaches log log n.

*/

#include <iostream>

using namespace std;

//Finding Prime Numbers using the sieve of Eratosthenes, one of a number of prime number sieves,

//is a simple, ancient algorithm for finding all prime numbers up to any given limit.

// it is not an trial and check algorithm but find prime numbers by jumping out.

//The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so)

int find_prime (int *numArray, int maxNum)

{

int factor = 2; // we will make 2 as the starting point.

numArray[1] = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.

//maxNum is the number till there we can find the prime numbers.

while ((factor * factor) < maxNum)

{

for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.

{

if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we haven’t yet checked it.

{

// we are putting zeros on all multiples of prime numbers, one by one.

if (numArray[i] != factor && (numArray[i]% factor) == 0)

{

numArray[i] = 0;

}

}

}

++factor;

}// after the loop, array will have zeros on all non prime locations and prime numbers.

}

int main ()

{

int maxNum = 0;

int result = 1;

int num = 0;

cout << “enter max number: “;

cin >> maxNum;

int *myArray = new int [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.

for (int i = 0; i < maxNum; i++)

{

myArray[i]= i;

}

// we will get prime numbers till the maxNum by calling below function.

find_prime(myArray, maxNum);

// returning array will have prime numbers and zeros on non prime locations.

// we start our loop from 2 as first two locations are not use full so we reduce the itration of the loop.

for (int i = 2; i < maxNum; i ++)

{

// we only do our calculation on prime numbers.

if ((myArray [i]) != 0)

{

num = myArray[i];

// we take multiples of each prime number till the maxNum. meaning we raise each prime to power of a number.

// which result will remain till maxNum.

while ((num * (myArray[i])) < maxNum)

{

num = num * (myArray[i]);

}

result = num * result;

}

}// after the loop result will have smallest positive number that is evenly divisible by all of the numbers from 1 to maxNum.

cout << “smallest positive number that is evenly divisible by all of the numbers from 1 to ” << maxNum

<< ” is: ” << result <<endl;

delete []myArray;

}

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