#include using namespace std; //The normal power funktion to solve b to the power of e(b^e) int pow(int b,int e) //gets base and exponent(b,e) { int res=b; //sets the result to base while(--e) //goes through e-1 times cause positive Integers are true and 0 is false res*=b; //multiplies res all e-1 times by b return res; //so the returned res is b^e } //This function gets to arrays with numbers and calculates how many numbers are correct but maybe not at the correct place, so '132' and '234' would result in two, because '2' and '3' are in both //This Function is not perfect! For '11' and '11' it gives 4 instead of 2, because the first one of the first array can be found twice in the second array and the second one of the first array can be found twice in the second array too, so there are 4 equal Numbers //This could be improved but then the function would need more time... If anyone wants me to improve this I can do int correctNumbers(int* correct, int* other, int length) //gets two integer-pointers pointing at an array and an integer which is supposed to give the length of both arrays(have to have the same length) { int res=0; //sets the result to 0 for(int i=0;i=0;j--) //Here we go through every digit of our case, starting at the end { cases[i][j]=temp%10; // we set the digit where we are to the last digit of temp temp/=10; //and then we move this digit away } //For example if we have temp as 193, we take 3 at first and then we divide by ten to get 19, then we take the 9 and the next round the 1 } for(int j=0;j

Crack the Code