POJ1047解题报告
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题目如下:
Round and Round We Go
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8338 Accepted: 3770
Description
A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a"cycle"of the digits of the original number. That is, if you consider the number after the last digit to "wrap around"back to the first digit, the sequence of digits in both numbers will be the same, though they may start at different positions.For example, the number 142857 is cyclic, as illustrated by the following table:
142857 *1 = 142857
142857 *2 = 285714
142857 *3 = 428571
142857 *4 = 571428
142857 *5 = 714285
142857 *6 = 857142
142857 *1 = 142857
142857 *2 = 285714
142857 *3 = 428571
142857 *4 = 571428
142857 *5 = 714285
142857 *6 = 857142
Input
Write a program which will determine whether or not numbers are cyclic. The input file is a list of integers from 2 to 60 digits in length. (Note that preceding zeros should not be removed, they are considered part of the number and count in determining n. Thus, "01"is a two-digit number, distinct from "1" which is a one-digit number.)
Output
For each input integer, write a line in the output indicating whether or not it is cyclic.
Sample Input
142857142856142858010588235294117647
Sample Output
142857 is cyclic142856 is not cyclic142858 is not cyclic01 is not cyclic0588235294117647 is cyclic主要考察的知识点就是大整数的乘法。我这一篇是前面一篇《判断一个数是否是循环数的算法》的后续,当时写那篇文章时对于大整数问题还不是很熟,所以
没办法解决数的溢出的问题。现在了解到,大整数的解决办法就是利用数组。只要知道了这个思想,就好办了。写代码细心一点,就能够AC。
代码如下:
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