Multiplying by Rotation——进制循环节

来源:互联网 发布:java web即时通讯源码 编辑:程序博客网 时间:2023/02/03 01:00

  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



个人认为这个题目很有意思,虽然看了很长时间也没有看懂,当理解了之后,瞬间感觉有趣得很。

题意大致是这样的:给出三个数,第一个数代表了进制(x),第二个数代表了要乘的因式的个位数(y),第三个数代表乘数(z),求一个数a,乘以第三个数(z)之后的最高位等于a的最低位(y)的位数。

现在,我们不知道a是多少,只知道a的最低位是y。不过很巧合的是,我们可以将这个数乘出来,即不断地将两个最低位相乘,直到进位为y,在这个过程中,记录相乘的次数,这样就得出了所需要的位数

比如说已知10,7,4

我们要找的是a乘以4之后的最高位是7的数。

我们先用7 * 4 = 28,留下一个2,进上一个8,最低位变成8;

用8 * 4 + 2 = 34,留下一个3,进上一个4,最低位变成4;

用4 * 4 + 3 = 19,留下一个1,进上一个9,最低位变成9;

用9 * 4 + 1 = 37,留下一个3,进上一个7,最低位变成7;

用7 * 4 + 3 = 31,留下一个3,进上一个1,最低位变成1;

用1 * 4 + 3 = 7;留下一个0,进上一个7,最低位变成7;

接下来,是7 * 4 + 0 = 28;这一步,和第一步重合,接下来就会陷入循环,我们找到的最低位依次是:7,8,4,9,7,1,从高位到低位连接起来就是:179487,179487 * 4 = 717948。



再描述一个17进制的例子:17,14,12

用14 * 12 = 168(10) = 9,15(17)(9 * 17 + 15 = 168)

用15 * 12 + 9 = 189(10) = 11,2(17)(11 * 17 + 2 = 189)

用2 * 12 + 11 = 35(10) = 2,1(17) (2 * 17 + 1)

用1 * 12 + 2 = 14(10) = 0,14(17) (0 * 17 + 14)

接下来就是循环,所以只需要4位就能找到这个数。


注意:如果乘以的因数为1,则什么数乘以这个数都是那个数本身,所以只需要一步

如果要乘的数的最低位为0,则乘任何数都是它本身,所以也只需要一步,这两种特殊情况要单独处理哦。


#include <stdio.h>#include <string.h>int main(){    int x,y,z,f;    int i,j;    while(~scanf("%d%d%d",&x,&y,&z))    {        i = 1;        j = z * y;        while(1)        {            i++;            j = j % x * z + j / x;            if(j == y)                break;        }        if(z == 1 || y == 0)            i = 1;        printf("%d\n",i);    }    return 0;}



0 0
原创粉丝点击