poj 1047 Round and Round We Go 字符串的处理

Round and Round We Go

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10716		Accepted: 4913

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

post code：原字符串赋值两次

然后 在每次生成的新的字符串中进行查找，如果木有找到 就跳出 返回失败

首先先 将

#include<stdio.h>
#include<string.h>
char temp[100];
char sample[100];
char ch[200];
int  num[100],len;
int  add()   //add 使字符串相加
{
     int i,carry=0;
     for(i=len-1;i>=0;i--)
     {
        temp[i]+=num[i]+carry;
        if(temp[i]>'9'){
                       carry=1;                                 
                       temp[i]-=10;
                       }
        else carry=0;
     }   
     return carry;  
}

int main() 
{
     int i,judge;
     char *pos;
     while(scanf("%s",temp)!=EOF)
     {
          
          len=strlen(temp);
          strcpy(ch,temp);   // ch 进行判断 将原字符串链接到后面
          strcat(ch,temp);
          strcpy(sample,temp);  //sample 保存原字符
          for(i=0;i<len;i++)
          {
            num[i]=temp[i]-'0';                 
          }
          for(i=1;i<len;i++)
          {
             judge=add();          //进行叠加 *1就可以忽略了 一定可以
             if(judge==1)break;
             else {
                     pos=NULL;
                     pos=strstr(ch,temp);     //先查找 如果找到就可以了
                     if(pos==NULL)break;         
                  }                  
          }
          if(i==len)printf("%s is cyclic\n",sample);    //进行了n次 每次都能找到就可以了
          else printf("%s is not cyclic\n",sample);     //木有找到
                                             
     }    
}