UVa 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

 

刚开始，我题意也不是很清楚，后来看了网上的大牛的解题报告，懂了。

题意：给出一个进制，一个数的最低位，和另外的一个数，比如10进制，第一个数字的最低位是7，第二个数字是4，

和规则（XXXXX7 * 4 = 7XXXXX，例子: 179487 * 4 = 717948 ）求出第一个数字的最小长度。

看起来很难，其实动笔写写就明白了。

输入k，m，n，原来的数字为s，因为s的最后一位为m，则s*n的最后一位为s*n%k，而s*n%k又是s的倒数第二位，这样又可以计算出ans*n的倒数第二位；

以此类推，直到乘积+原来的进位==最低位。

AC代码：

#include<stdio.h>int main(){int k,m,n,c,s;while(scanf("%d%d%d",&k,&m,&n)!=EOF){s=m*n;c=1;while(s!=m){s=s%k*n+s/k;c++;}printf("%d\n",c);}return 0;}