Implement large number multiplication using a string or an array.

Approach

Create an array called largeInteger, which represents a large number.

Fill the largeInteger with n.

Iterate over numbers from n to 0:

Multiply largeInteger by this number.

Handle the potential carry.

Output the content of largeInteger to stdout.

Pseudocode of largeIntMultiply:

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largeIntMultiply(largeInt, n): Input largeInt a very large number represent by an array n the multiplier Output the largeInt multiplied by n carry = 0 for every i from largeInt.size()-1 to 0: number = n * largeInt[i] + carry largeInt[i] = number % 10 carry = number / 10 if i == 0 and carry != 0: add a 0 at the beginning of largeInt i++ end if end for return largeInt

Pseudocode of extraLongFactorials:

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extraLongFactorials(n): Input n the number to be calculated largeInteger = empty array i = n for i != 0: add i % 10 to the beginning of largeInteger i /= 10 end for for every i from n-1 to 2: largeInteger = largeIntMultiply(largeInteger, i) end for output largeInteger to stdout

voidlargeIntMultiply(vector<int> &largeInt, int n); voidextraLongFactorials(int n){ // Convert int to vercor<int> vector<int> largeInteger; for (int i = n; i != 0; i /= 10) { largeInteger.insert(largeInteger.begin(), i % 10); } for (int i = n-1; i >= 2; i--) { largeIntMultiply(largeInteger, i); } for (int i : largeInteger) { cout << i; } }

voidlargeIntMultiply(vector<int> &largeInt, int n){ int carry = 0; for (int i = largeInt.size()-1; i >= 0; i--) { int number = largeInt[i] * n + carry; largeInt[i] = number % 10; carry = number / 10; if (i == 0 && carry != 0) { largeInt.insert(largeInt.begin(), 0); i++; } } }