Project Euler Problem Solved

The following iterative sequence is defined for the set of positive integers: n → n /2 ( n is even) n → 3 n + 1 ( n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? #include <bits/stdc++.h> using namespace std; bool max(int &last_max,int count,int &max_num,int i) {      if(count > last_max)      {           last_max =count;         max_num = i;      }   } int main() {   int count = 0;    int last_max = -999;   int max_num = 0;    for(long l...

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. #include <bits/stdc++.h> using namespace std; int main() {   long long int n=100;   long long sum1 = (n*(n+1))/2;   long long sum2 = (n*(n+1)*(2*n+1))/6;   long long int diff = (sum1*sum1) - sum2;   cout << diff <<endl;    }

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. #include <bits/stdc++.h> using namespace std; int main() {         int sieve[2000000];     sieve[0] = 0;     sieve[1] = 0;        for(int k=2; k < 2000000; k++)        {            sieve[k] = 1;        }     for(int i = 2;i<2000000;i++)     {         if(sieve[i]==1)         {             for(int j = 2*i;j<2000000;j+=i)             {                 sieve[j] = 0;             }  ...

A Pythagorean triplet is a set of three natural numbers, a < b < c , for which, a 2 + b 2 = c 2 For example, 3 2 + 4 2 = 9 + 16 = 25 = 5 2 . There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc . #include <bits/stdc++.h> using namespace std; int main() {   for(int i=0; i< 1000; i++)   {     for(int j=i+1; j<1000; j++)     {       for(int k=j+1; k<1000; k++)       {          if(i + j + k == 1000 && (i*i + j*j == k*k))            cout << i*j*k << endl;       }     }   } }

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 845801...

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? #include <bits/stdc++.h> using namespace std; int main() {         int sieve[1000001];     sieve[0] = 0;     sieve[1] = 0;        for(int k=2; k <= 1000000; k++)        {            sieve[k] = 1;        }     for(int i = 2;i<=1000000;i++)     {         if(sieve[i]==1)         {             for(int j = 2*i;j<=1000000;j+=i)             {                 sieve[j] = 0;       ...