2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13345    Accepted Submission(s): 4146

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

 

#include<cstdio>
#include<cmath>
int Powermod(int a,int b,int c)//快速幂
{
    int ans=1;
    if(a%c==0) return 0;
    a=a%c;
    while(b)
    {
        if(b&1)
            ans=ans*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ans;

}
int main()
{
    int i,n;
    //奇数除了1一定有结果，偶数一定没结果
    while(~scanf("%d",&n))
    {
        if(n%2==0||n==1)//2^x对偶数求余结果为偶数，不为1   1的时候结果也不存在
       {printf("2^? mod %d = 1
",n);continue;}
            for(i=1;; i++)//对于2^x mod n，当1<=i<=n 就能得到所有求余结果
            if(Powermod(2,i,n)==1)
            {
                printf("2^%d mod %d = 1
",i,n);
                break;
            }

    }
}