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Project Euler Problem#11 solution in C++

In the 20×20 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 20×20 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];
}
}

Project Euler problem#9 solution in C++

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Solution:

#include <iostream>
using namespace std;

int main()
{
int a;
int b;
int c;
for (a=1; a<=500; a++)
{
for (b=1; b<=500; b++)
{
c=1000-b-a;
if (a*a+b*b-c*c == 0 && a<b )
{
cout<<a<<” “<<b<<” “<<” “<<c<<” “<<a*b*c<<endl;
}
}
}
return 0;
}

Project Euler Problem# 8 solution in C++

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

Solution:

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
int maxNum =1;
char a ;
string test ;
int result =1;
int checker = 1;

string numString  = (“7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450”);

for (int i = 0; i < 1000; i++)
{
if (numString[i] == ‘9’)
{
for (int j = 0; j < 5; j++)
{
a = numString[i+j];
maxNum *= atoi(&a);
}
if (result < maxNum)
{
result = maxNum;
}else
{
cout << result << endl;
}
maxNum = 1;
}
}
}

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;

}

Project Euler Problem# 6 solution in C++

The sum of the squares of the first ten natural numbers is,

12 + 22 + … + 102 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)2 = 552 = 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.

Efficient solution without using any loops. So it will be constant time method:

#include<iostream>

using namespace std;

int main ()
{
int maxNum = 0;
int sum_of_square = 0;
int square_of_sum = 0;

cout << “Enter Maximum number: “;
cin >> maxNum;

sum_of_square = ((maxNum*(maxNum +1))*((2*maxNum)+1))/6;

square_of_sum *= square_of_sum = maxNum* (maxNum+1)/2;

cout << “sum of squares till ” << maxNum << ” is: ” << sum_of_square << endl;
cout << “squares of sum till ” << maxNum << ” is: ” << square_of_sum << endl;
cout << “difference between the sum of the squares the square of the sum: ” << square_of_sum – sum_of_square  << endl;
}

Project Euler Problem #4 C++ solution:

/** (From:http://projecteuler.net/index.php?section=problems&id=3)

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.

Solution: (Brute force)

#include <iostream>
#include <string>
#include <sstream>

using namespace std;

void check(int);

int main ()
{
int result = 0;

for (int i = 999; i >= 101; i–)
{
for (int j = i; j >= 101; j–)
{
result = (j *i);
check (result);
}
}
}

void check (int a)
{
string str = static_cast<std::ostringstream*>( &(std::ostringstream() << a) )->str();
string syl_1;
string syl_2;
string syl_3;
static string j ;
int len = (str.length()/2);
syl_1 = str.substr(0, len);
syl_2 = str.substr(len);

for (int i = syl_2.length(); i >=0; i –)
{
if (syl_2[i] != NULL)
syl_3 += syl_2[i];
}
if ((syl_1.compare(syl_3))==0)
{
if (j < str)
j = str;
if (str == j)
cout << j << endl;
}
}

 

Project Euler: Problem # 3 solution in C++

/** (From:http://projecteuler.net/index.php?section=problems&id=3)

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

What is the largest prime factor of the number 600851475143 ? */

#include <iostream>

using namespace std;

int main ()
{
long long factorOf  = (600851475143);
int num = 2;

while ((num * num) <= factorOf)

if (factorOf % num == 0)
{
cout << num << endl;
factorOf /= num;
}else
num++;

cout << “factor of 600851475143 is: ” << factorOf << endl;
}

Project Euler: Problem #2 solution in C++

/**(From:http://projecteuler.net/index.php?section=problems&id=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.

Solution:*/

#include <iostream>

using namespace std;

int main()
{

int term_1 = 1;
int term_2 = 2;
int term_3 = 0;
int MaxTerm = 0;
int sum = 0;

cout<< “This Program lists the Fibonacci sequence. Please enter the Max number you want to have sequence:” << endl;

cin >> MaxTerm ;

while (term_1 <= MaxTerm)
{
cout<< term_1 << endl;
if ((term_1 %2) == 0){sum += term_1;}

term_3 = term_1 + term_2;
term_1 = term_2;
term_2 = term_3;

}
cout << “the sum of even value numbers is: “<< sum << endl;
}

Project Euler: Problem #1 solution in C++

/**(From:http://projecteuler.net/index.php?section=problems&id=1)

Problem Description:
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.

Solution:*/

#include <iostream>

using namespace std;

int main ()
{
int maxNum = 0;
int result = 0;

cout << “Please enter max number limit for the sum of all the multiples of 3 or 5: “;
cin >> maxNum;

for (int i = 2; i < maxNum; i++)
{
if (i % 5 == 0)
{
result += i;
}else if (i % 3 == 0)
{
result += i;
}
}
cout << “total of multiples of 3 or 5: ” << result << endl;

}