Description

　　　　　　　　　　　　　　　　　　　　Description

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 4
9 7 4
17 14 12

 

Sample Output 

6
2
4

 

题意：给出进制，末尾数字，乘数，求最小几位数可以满足原数乘完后等于原数的最后一位提前。例子：179487 * 4 = 717948

tip：

输入10 7 4

｛179487 * 4 = 717948（10进制）｝

4*7=28     28!=7    28/10=2    28=8;

4*8+2=34    34!=7    34/10=3    34=4;

4*4+3=19    19!=7    19/10=1    19=9;

4*9+1=37    37!=7    37/10=3    37=7;

4*7+3=31    31!=7    31/10=3    31=1;

4*1+3=7      7==7;(end)

输出6

 

#include<iostream>

using namespace std;

int main()
{
    int j,l,x;
    while(cin>>j>>l>>x)
    {
        int tmp=l*x,t1,t2,i=1;
        while(tmp!=l)
        {
            t1=tmp/j;
            t2=tmp%j;
            tmp=t2*x+t1;
            ++i;
        }
        cout<<i<<endl;
    }
    return 0;
}