Round and Round We Go
Time Limit:1000MSMemory Limit:10000K
Total Submissions:12064Accepted:5630
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 = 142857142857 *2 = 285714142857 *3 = 428571142857 *4 = 571428142857 *5 = 714285142857 *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
题意：给出一个字符串，如果这个字符串分别用1,2,…,len去乘，最终得到的结果还是由这个字符串的字符组成的，那么它是cyclic；
解题思路：给出的数*（len+1）=9…9（len个）,那么这个数是cyclic
参考代码：
#include <iostream>#include <string.h>using namespace std;char s[100];int a[100];int main(){while (cin>>s){int len=strlen(s);memset(a,0,sizeof(a));int k=0,left=0;for (int i=len-1;i>=0;i–){int ans=(s[i]-'0')*(len+1)+left;a[k++]=ans%10;left=ans/10;}while (left!=0){a[k++]=left%10;left/=10;}int flag=0;for (int i=0;i<k;i++){if (a[i]!=9){flag=1;break;}}if (flag==1)cout<<s<<" is not cyclic"<<endl;elsecout<<s<<" is cyclic"<<endl;}return 0;}
，就看你能不能战胜它。战胜了，你就是英雄，就是生活的强者。
