UVA550 UVALive5566 Multiplying by Rotation【进制+迭代】

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

问题链接：UVA550 UVALive5566 Multiplying by Rotation
问题简述：
????一个数乘以基数，等于原来数的最低位移到最高位。求这样的数最多有几位？给定基数、尾数（最低位）和因子。
问题分析：
????进制问题，通过迭代计算解决。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA550 UVALive5566 Multiplying by Rotation */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int base, last, factor;
    while(scanf("%d%d%d", &base, &last, &factor)!=EOF) {
        int cnt = 1, tmp = factor * last;
        while(last != tmp) {
            tmp = tmp % base * factor + tmp / base;
            cnt++;
        }

        printf("%d\n", cnt);
    }

    return 0;
}

UVA550 UVALive5566 Multiplying by Rotation【进制+迭代】

原文：https://www.cnblogs.com/tigerisland45/p/10408918.html