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

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12146 Accepted Submission(s):
3797

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

Sample Output

2^? mod 2 = 1

2^4 mod 5 = 1

题目大意：暴力搜索，找到合适的X值，这一题可以采取反过来暴力寻找，这一简单易懂些。

要注意的是输出的值时都要变化的，输出注意一下就好了，毕竟我是wa过的。。。

 #include <iostream>

 #include <cstdio>

 using namespace std;

 int main ()

 {

     int n;

     while (cin>>n)

     {

         if (n%&&n>)

         {

             int s=,x=;

             while (x)

             {

                 s=s*%n;

                 if (s==)

                 {

                     printf ("2^%d mod %d = 1\n",x,n);

                     break;

                 }

                 x++;

             }

         }

         else

             printf ("2^? mod %d = 1\n",n);

     }

     return ;

 }

巴特西

hdu 1395 2^x mod n = 1（暴力题）

2^x mod n = 1

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