2^x mod n = 1 HDU1395

/*2^x mod n = 1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6738    Accepted Submission(s): 2022

Problem Description Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

Input One positive integer on each line, the value of n.

Output If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.

Sample Input 2 5

Sample Output 2^? mod 2 = 1 2^4 mod 5 = 1

Author MA, Xiao

Source ZOJ Monthly, February 2003 */

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
bool h[10000];
int main()
{
    int n,k,n1,l,c;
    while(scanf("%d",&n)!=EOF)
    {
        memset(h,0,sizeof(h));
        k=0;l=1;c=2;
       printf("2^");
       if(n==1||n==2)
       {
           printf("? mod ");
           printf("%d = 1\n",n);
           continue;
       }
       n1=n;
       while(c%n1!=1)
       {
           h[c]=1;
           c*=2;
           c=c%n1;
           l++;
           if(c!=1)
           {
               if(h[c]==1)//经典的比较方法，与前面的c比较，值得收藏。
               {
                   k=1;
                   break;
               }
               else
               h[c]=1;
           }
       }
       if(k==1)
       {
           printf("? mod ");
           printf("%d = 1\n",n);
           continue;
       }
       else
       {

           printf("%d mod ",l);
           printf("%d = 1\n",n);
       }
    }
    return 0;
}//本题的为偶数时一定不到l，因为2^l=n*q+r.......r一定为偶数不会是1，所以可以把偶数单独拿出来。