HDU 3307 description has only two sentences

 

Description has only two sentences

Time Limit: 3000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 886 accepted submission (s): 265

Problem descriptionan = x * an-1 + Y and y MOD (X-1) = 0.
Your task is to calculate the smallest positive integer k that AK mod a0 = 0.

 

Inputeach line will contain only three integers x, y, a0 (1 <x <231, 0 <= Y <263, 0 <A0 <231 ).

 

Outputfor each case, output the answer in one line, if there is no such K, output "impossible! ".

 

Sample input2 0 9

 

Sample output1

 

Authorwhereisherofrom

 

Sourcehdoj monthly contest-2010.02.06

 

Ask for a minimum K of AI = Xai + Y and Y % (x-1) = 0 AK % A0 = 0

Then start to push the formula

A n = x ^ N * A0 + (x ^ (n-1) + x * (n-2) +... + x + 1) * y

Obviously, x ^ N * a0 must be divisible by A0, which can be ignored, and then the following item is an proportional sequence.

Then, the minimum k (x ^ k-1)/(x-1) * Y % A0 = 0 is obtained.

.. Set the coefficient to E,

E * a0 = (x ^ k-1)/(x-1) * Y;

E * (A0/gcd (A0, Y/(x-1) = x ^ k-1.

Set M = (A0/gcd (A0, Y/(x-1 ))).

Obtain x ^ K = 1 (mod m)

 

Evaluate a single Euler for M and judge it.

Impossible should determine whether X and m are mutually qualitative. If not, x ^ K % m must be a multiple of gcd (M, X ).

 

#include <algorithm>#include <cstring>#include <cstdio>#include <queue>#include <iostream>#include <vector>#include <string>#include <map>using namespace std;typedef long long ll;const int N = 100010;ll mod , tot ;ll gcd ( ll a , ll b ) { return b == 0 ? a : gcd( b ,a % b ); }ll eular ( ll n ){    ll sum = n ;    for( ll i = 2 ; i* i <= n ; ++i ){        if(  n % i == 0 ){            sum = sum / i *( i -1 );            while( n % i == 0 ) n /= i ;        }    }    if( n > 1 ) sum = sum / n * ( n -1 ) ;    return sum;}ll quick_mod( ll a , ll b ){    ll res = 1 ;    while( b ){        if( b & 1 ) res =  res * a % mod ;        a = a * a % mod;        b >>= 1 ;    }    return res ;}ll e[N] ;void get_factor( ll n ){    for( ll i = 2 ; i * i <= n ; ++i ){        if( n % i == 0 ){            e[tot++] = i ;            if( n / i != i ) e[tot++ ] = n / i ;        }    }}int  main(){    ios::sync_with_stdio(0);    #ifdef LOCAL        freopen("in.txt","r",stdin);    #endif    ll a0 , x , y ;    while( cin >> x >> y >> a0 ){        tot = 0 ;        if( !y ) { cout << ‘1‘ <<endl ; continue ;}        ll m = a0 / gcd( y / ( x - 1 )  ,a0 ) ;        if( gcd(m ,x) != 1 ) { cout << "Impossible!" << endl ; continue ; }        ll phi = eular(m) ;        get_factor(phi) ;        sort(e , e + tot ) ;        mod = m ;        for( int i = 0 ; i < tot ; ++i ){            if( quick_mod( x,e[i]) == 1 ){ cout <<e[i] << endl ; break ; }        }    }}

 

HDU 3307 description has only two sentences