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

 

 

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#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",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\n",i,n);                 break;             }       } }