杭州电 1395

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<m。设x为欧拉函数的值，则a^x%m=1恒成立。因为题上的说明是a为二

则仅仅要m是奇数，且m不等于1就可以

则有一下代码：

 #include<stdio.h>
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);
 }
}

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