UVa550 Multiplying by Rotation

最新推荐文章于 2016-07-11 16:37:54 发布

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最新推荐文章于 2016-07-11 16:37:54 发布

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本文介绍了一种特殊的乘法运算，在特定的进制下，某个自然数与其因子相乘后，其末位数字能够旋转至首位，文章提供了一个算法实现来找出具有此特性的最小数。

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Multiplying by Rotation

Warning: Not all numbers in this problem are decimal numbers!

Multiplication of natural numbers in general is a cumbersome operation. In somecases 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 aboveexample 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

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 standardend-of-file marker.

Output

Your program determines for each input line the number of digits of the smallest first factor withthe rotamultproperty. The output-file is also a textfile. Each line contains the answer for thecorresponding input line.

Sample Input

10 7 4
9 7 4
17 14 12

Sample Output

6
2
4

Miguel A. Revilla
1998-03-10

这道题目看起来很复杂，但要理解了就会发现只要套公式循环即可，题目是在n进制下，给予一个乘数因子和一个尾数。要求求出某数能在该进制下通过乘法将该尾数从此数的尾部移动到首部，其他数字不变，通过tmp % system * factor + tmp / system循环，模拟乘法过程求出首部，不断判断首部是否等于尾部，并在每次循环中使某数长度+1，最后输出长度。

#include <iostream>
using namespace std;

int main() {
	int system,last,factor;

	while (cin >> system >> last >> factor) {
		int tmp = factor * last;
		int cnt = 1;
		while (1) {
			if (last == tmp)
				break;
			tmp = tmp % system * factor + tmp / system;
			cnt++;
		}
		cout << cnt << endl;
	}
	return 0;
}