hdu 1395 2^x mod n = 1（暴力题）

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

2

5

 

 

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 ;
 }