550 - Multiplying by Rotation

  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

题意：承认自己英文太差。自己没有完全看多，看了别人的翻译才懂的题意。

大致就是给你一串数，是k进制的，然后乘以给定的一个chen之后，得到新的一个数，当然也是k进制的。那么这个乘积正好可以有原先的数去掉最后一位，然后放到最前面一位而构成

理解题意之后，其实就是一个模拟乘法的过程。比较简单

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>using namespace std;int main (){    int base,ge,chen,sum,a,b;    while(cin>>base>>ge>>chen)    {        sum=0;        a=ge;        b=0;        while(1)        {            int t;            t=a*chen+b;            a=t%base;            b=t/base;            sum++;            if (a==ge && b==0) break;        }        cout<<sum<<endl;    }    return 0;}