2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11658    Accepted Submission(s): 3634


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       思路就是：欧拉定理 就是a和m互质，且a
int main()
{
 int n;
 while(~scanf("%d",&n))
 {
  if(n%2&&n>1)
  {
   int i,s=1;
   for(i=1;;i++)//由于推断了奇数一定有解 所以能够用这种暴力 由于一定能跳出
   {
    s=s*2%n;
    if(s==1)
    {
     printf("2^%d mod %d = 1 ",i,n);
     break;
    }
   }
  }
  else
  printf("2^? mod %d = 1 ",n);
 }
} 版权声明：本文博主原创文章。博客，未经同意不得转载。