Hdu1395-2 ^ x mod n = 1

The principle is a ^ B % n = d; >>>>>> (a % n) × ........ * (a % n) % n = d

Then when n = 1 or n % 2 = 0, d must be 0, so there is no solution at this time;

When n is another value, it must be 1 ~ The remainder of n-1 exists. Therefore, use the method for solving a ^ B % n = d.

 

#include<iostream>   #include<cstdio>   #include<cstring>   #include<cmath>   #include<algorithm>   #include<bitset>   #include<iomanip>     using namespace std;    int main()  {      int a , b , n ;      while( ~scanf( "%d" ,&n )  )      {          if( n == 1 || n % 2 == 0 )          {              printf( "2^? mod %d = 1\n" , n );          }          else          {              b = 1 ;              int temp = 2 ;               while( temp != 1 )              {                  b++ ;                  temp = ( temp * 2 ) % n ;              }              printf( "2^%d mod %d = 1\n" , b , n ) ;          }      }         return 0 ;  }  #include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<bitset>#include<iomanip>using namespace std;int main(){int a , b , n ;while( ~scanf( "%d" ,&n )  ){if( n == 1 || n % 2 == 0 ){printf( "2^? mod %d = 1\n" , n );}else{b = 1 ;int temp = 2 ; while( temp != 1 ){b++ ;temp = ( temp * 2 ) % n ;}printf( "2^%d mod %d = 1\n" , b , n ) ;}}return 0 ;}