uva 550 Multiplying by Rotation（模拟）

  Multiplying by Rotation 

Warning: Not all numbers in this problem are decimal numbers!

Multiplication of natural numbers in general is a cumbersome operation. In some cases however the product can be obtained by moving the last digit to the front.

Example: 179487 * 4 = 717948

Of course this property depends on the numbersystem you use, in the above example we used the decimal representation. In base 9 we have a shorter example:

17 * 4 = 71 (base 9)

as (9 * 1 + 7) * 4 = 7 * 9 + 1

Input 

The input for your program is a textfile. Each line consists of three numbers separated by a space: the base of the number system, the least significant digit of the first factor, and the second factor. This second factor is one digit only hence less than the base. The input file ends with the standard end-of-file marker.

Output 

Your program determines for each input line the number of digits of the smallest first factor with the rotamultproperty. The output-file is also a textfile. Each line contains the answer for the corresponding input line.

Sample Input 

10 7 49 7 417 14 12

Sample Output 

624

题目大意：给出进制bas，数b, 乘数c。求一个数bas进制的num,num的要求是各位必须为b，把个位数b放到num的首位形成新的一个数sum，然后sum要等于num * c，输出num的位数。

解题思路：题目很有意思，因为num = tem.b;而sum=b.tem;也就是说，（当前位*c + 进位）模去 bas =当前位的前一位。

#include<iostream>#include<math.h>using namespace std;int main(){int bas, b, c;while (cin >> bas >> b >> c){int cnt = 0, p = 0, sum, over = b;while (1){cnt++;sum = b * c + p;b = sum % bas;p = sum / bas;if (over == b && p == 0)break;}cout << cnt << endl;}return 0;}