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%2&&n>1)
        {
            int s=1,x=1;
            while (x)
            {
                s=s*2%n;
                if (s==1)
                {
                    printf ("2^%d mod %d = 1
",x,n);
                    break;
                }
                x++;
            }
        }
        else
            printf ("2^? mod %d = 1
",n);
    }
    return 0;
}