Round and Round We Go

最新推荐文章于 2024-07-31 17:49:42 发布

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最新推荐文章于 2024-07-31 17:49:42 发布

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本文介绍了一种用于检测循环数字的算法实现方法，通过字符串加法替代乘法操作，判断给定数字是否具备循环性质。

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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

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

142857
142856
142858
01
0588235294117647

Sample Output

142857 is cyclic
142856 is not cyclic
142858 is not cyclic
01 is not cyclic
0588235294117647 is cyclic

主要思想：输入的整数最大为60位，所以选择字符串接收。首先要解决的问题是怎么降低字符串乘法的复杂度，这里digit从1乘到N，所以可以用字符串加法代替乘法，通过N次循环将digit自增。接下来要解决的问题是如何判断当前得到的字符串是否为digit移位得到的，比较简单的方法就是将两个digit连接为exDigit，然后判断得到的字符串是否为exDigit的子串即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#define MAX 70
#include<algorithm>
using namespace std;
//实现dig1和dig2的字符串加法，结果更新dig1
//返回true运算成功，返回false说明得到的字符串长度大于len
bool AddSelfOnce(char dig1[],char dig2[],int len);
char digit[MAX];
char exDigit[MAX*2];
char temp[MAX];
int main()
{
    int i,len;
    while(cin>>digit)
    {
        len=strlen(digit);
        for(i=0; i<2*len; i++)        //初始化temp和exDigit
        {
            exDigit[i]=digit[i%len];
            if(i<len)temp[i]=digit[i];
            else if(i==len)temp[i]=NULL;
        }
        exDigit[i]=NULL;
        for(i=2; i<=len; i++)
        {
            if(!AddSelfOnce(temp,digit,len))
            {
                cout<<digit<<" is not cyclic"<<endl;
                break;
            }
            if(strstr(exDigit,temp)==NULL)    //如果temp不是exDigit的子串
            {
                cout<<digit<<" is not cyclic"<<endl;
                break;
            }
        }
        if(i>len)cout<<digit<<" is cyclic"<<endl;
    }
    return 0;
}
bool AddSelfOnce(char dig1[],char dig2[],int len)
{
    int i,carry=0;
    for(i=len-1; i>=0; i--)
    {
        dig1[i]+=(dig2[i]+carry-48);
        if(dig1[i]>'9')
        {
            if(0==i)
                return false;
            dig1[i]-=10;
            carry=1;
        }
        else carry=0;
    }
    return true;
}