思路：简单的模拟乘法
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int n,m,k;
while(scanf("%d%d%d",&m,&k,&n)!=EOF)
{
int count = 1,now=k,save=0,temp;
while(now*n+save!=k)
{
temp=now;
now = (temp*n+save)%m;
save = (temp*n+save)/m;
count++;
}
printf("%d\n",count);
}
}
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
