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。。。

 1 #include <iostream>

 2 #include <cstdio>

 3 using namespace std;

 4 

 5 int main ()

 6 {

 7     int n;

 8     while (cin>>n)

 9     {

10         if (n%2&&n>1)

11         {

12             int s=1,x=1;

13             while (x)

14             {

15                 s=s*2%n;

16                 if (s==1)

17                 {

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

19                     break;

20                 }

21                 x++;

22             }

23         }

24         else

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

26     }

27     return 0;

28 }