Uvaoj 550-multiplying by rotation

Question

Question omitted.

Analysis: I expanded the formula, pushed something out, enumerated digits, and then obtained an X, if X represents the number under the base where the current total number of digits minus one, the current number of digits is the minimum. After the write, TLE determines that it may take a long time.

After reading other people's questions, I found my consideration complicated. In fact, this is a multiplication operation of multiplying Multiple Digits by one digit in base bitwise. The vertical format is as follows,

A0a1a2... an-1c

× F

----------------------

Ca0a1a2... an-1

(A0a1a2... the an-1c represents factor1, f Represents factor2, and C represents the number to be moved ).

For example, after C * F is a bitwise operation, the result C * F is represented as C * F/base and (C * f) % base at base, where C * f/base is the carry, and C * F % base is the current bit of the result. It is determined by the question conditions, and an-1 = C * F % base.

So we continue with an-1 * F until the current bit of the result is equal to C and the carry value is 0. It can be seen that this is the type of question given by elementary school students to leave a few blank spaces, but it is changed from decimal to base.

Code:

 
# Include <iostream> # include <cmath> # include <string> using namespace STD; int main () {int base, n, m; while (CIN> base> N> m) {If (M = 1) {cout <1 <Endl; continue;} int carry = 0, cur = N, I = 1; for (; I ++) {int TMP = cur * m + carry; carry = tmp/base; // enter the full base into the cur = TMP % base; // current bit if (carry = 0 & cur = N) {cout <I <Endl; break; }}return 0 ;}
 

Tle code:

# Include <iostream> using namespace STD; int B, C, F; Int Is (INT X, int N) // can X be represented as the N-digit number in B {int CNT = 0; For (; X/= B, CNT ++); If (CNT = N) return 1; else return 0;} int main () {While (CIN> B> C> F) {int I; // number of digits of the first factor n int B _pow = 1; // n-1 times of B for (I = 1; B _pow <= f; I ++) B _pow * = B; // ① can n be 1? ② <=, X cannot be 0? For (; I ++, B _pow * = B) {int T1 = C * (B _pow-f); int t2 = (F * b-1); If (T1% T2! = 0) continue; // X is an integer int x = t1/T2; If (is (x, I-1) {cout <I <Endl; break ;}}}}