hdu 数论专题

hdu 1359

2^x mod n = 1

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

分析：

当n==1 || n为偶数时 ，2^x mod n != 1 ; 故 n>1 且 n 为奇数时 ，从2开始遍历能找到x ，使等式成立。

2^x % n = (2^a %n) *(2^b%n)%n;  其中 x= a+b.

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<map>
#include<math.h>

using namespace std;

int main()
{
    int n;
    while(cin>>n)
    {
        int mi,tmp;
        if(n==1 || n%2==0)
            cout<<"2^? mod "<<n<<" = 1"<<endl;
        else
        {
            mi=1;
            tmp=2;
            while(tmp!=1)
            {
                tmp=tmp*2%n;
                mi++;
            }
            cout<<"2^"<<mi<<" mod "<<n<<" = 1"<<endl;
        }

    }

    return 0 ;
}