How to Approach?

#1

Please guide on how to approach the problem.

#2

@Ak07 we cant store result in long long int so we will store digits of the resulted value in an array,
do the following:
Create an array ‘res[ ]’ of MAX size where MAX is number of maximum digits in output.
Initialize value stored in ‘res[ ]’ as 1 and initialize ‘res_size’ (size of ‘res[ ]’) as 1.
Do following for all numbers from x = 2 to n……

a) Multiply x with res[ ] and update res[ ] and res_size to store the multiplication result.

multiply(res[ ], x)

Initialize carry as 0.
Do following for i = 0 to res_size – 1 ……

a) Find value of res[i] * x + carry. Let this value be prod.

b) Update res[i] by storing last digit of prod in it.

c) Update carry by storing remaining digits in carrying.

Put all digits of carry in res[ ] and increase res_size by number of digits in carry.

#3

I don’t understand the logic behind this method, or is this something we just need to know and memorise?

#4

@Ak07 here we are using basic multiplication technique which we studied in primary school.look at this code you will understand (https://ide.codingblocks.com/s/195158)

#5

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