Ultraviolet A 550: multiplying by rotation

This question can be pushed forward to the number of all BITs by the known last bits of the first multiplier.

For example:

The given base number is 10, the last digit of the first multiplier is 7, and the second multiplier is 4.

List the following formula:

7

X 4

-------------------------- The carry value is 0.

Next, you will get:

87

X 4

-------------------------- Carry 2

8

Next step:

487

X 4

-------------------------- The carry value is 3.

48

Next step:

9487

X 4

-------------------------- Carry is 1

948

Next step:

79487

X 4

-------------------------- The carry value is 3.

7948

Next step:

179487

X 4

-------------------------- The carry value is 3.

17948

Next step:

179487

X 4

-------------------------- The carry value is 3.

717948

So far, the first multiplier is obtained.

This question can be simulated in this way to obtain the result.

My code is as follows:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;int main(){int base,d,f;// base, last significant digit of the first factor, the second factorwhile(cin >> base >> d >> f){int count = 0;//count for the first factor's digits numberint product, carry, next_digit;//product of next_digit and f, carry of product baseed on base, next digit of the fist factornext_digit = d;carry = 0;while(1){count++;product = next_digit*f;if (product+carry == d){break;}next_digit = (product+carry)%base;carry = (product+carry)/base;}cout << count << endl;}return 0;}