HDU-1395 2^x mod n = 1

http://acm.hdu.edu.cn/showproblem.php?pid=1395

怎样取余是关键。。

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9534    Accepted Submission(s): 2932

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

 

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 #include<stdio.h>
 int main()
 {
     int n,ans,s;
     while(scanf("%d",&n)!=EOF)
     {
         if(n%==||n==)
             printf("2^? mod %d = 1\n",n);
         else
         {
             ans=;s=;
             while(s!=)
             {
                 ans++;
                 s=s*%n;//取余。。。
             }
             printf("2^%d mod %d = 1\n",ans,n);
         }
     }
     return ;
 }