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 Description

a_n = X*a_n-1 + Y and Y mod (X-1) = 0.
Your task is to calculate the smallest positive integer k that a_k mod a₀ = 0.

 

Input

Each line will contain only three integers X, Y, a₀ ( 1 < X < 2³¹, 0 <= Y < 2⁶³, 0 < a₀ < 2³¹).

 

Output

For each case, output the answer in one line, if there is no such k, output "Impossible!".

 

Sample Input

2 0 9

 

Sample Output

1

 

Author

WhereIsHeroFrom

 

Source

HDOJ Monthly Contest – 2010.02.06

题目要求求一个  Ai =  XAi + Y and Y %(x-1) == 0 的  Ak % A0 == 0 的最小 k

然后开始推公式

可得   A n  = X^n *  A0  + (  x^(n-1) + x *(n-2) + ...... + x + 1 ) * Y

显然  X^n * A0 必然整除 A0 可以忽略  ， 然后  后项是一个等比数列 。

然后处理一下得到想要求的就是   （ X ^ k -1 ）/ ( x - 1 ) * Y    %  A0  ==  0 的最小k

转换一下    。。   设系数为  e  ,

就是    e * A0  = ( X^k - 1 ) / (X - 1 )  * Y ;

          e * ( A0 / gcd(A0 , Y / (X - 1 ) ) ) =  X^k - 1 。 

          设  m  = ( A0 / gcd(A0 , Y / (X - 1 ) ) )  . 

          得到   X^k == 1 ( mod m )

对m求一次单个欧拉 ， 然后判一下 。 

impossible 那里要特判一下X与m是否互质  ，若不互质，则X^k%m必定是gcd(m,X)的倍数  

#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 ; }
        }
    }
}