Description has only two Sentences hdu-3307（欧拉定理）

Description has only two Sentences

 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 a 0 = 0.

Input
Each line will contain only three integers X, Y, a 0 ( 1 < X < 2 31, 0 <= Y < 2 63, 0 < a 0 < 2 31).
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

题意：

题目给你三个数$X,Y,a_0$我们知道了一个等式，$a_n=X*a_{n-1}+Y$并且$Y\ mod \ (X-1) ==0$ 让我们求一个最小的k使得$a_k\ \% \ a_0 =0$

分析：

我们先把每个式子写出了观察一下规律

$a_0 = a_0$

$a_1 = Xa_0+Y$

$a_2 = Xa_1+Y = X(Xa_0+Y)+Y = X^2a_0+XY+Y$

$a_3 = Xa_2+Y = X(X^2a_0+XY+Y)+Y = X^3a_0+X^2Y+XY+Y$
.
.
.
$a_n = Xa_{n-1}+Y = X^na_0+X^{n-1}Y+X^{n-2}Y+\cdots+XY+Y$
$\ \ \ \ =X^na_0 + Y(1+X+X^2+X^3+\cdots+X^{n-1})$
$\ \ \ \ =X^na_0 + Y\cdot \frac{X^n-1}{X-1}$

因此我们发现了通项公式

$a_k = X^ka_0 + Y\cdot \frac{X^k-1}{X-1}$

所以

$a_k\ \%\ a_0 = Y\cdot \frac{X^k-1}{X-1}\ \%\ a_0 = 0$

因此我们得到

$a_0\ |\ \frac{Y}{X-1}\cdot (X^k-1)$

我们设$d = gcd(a_0,\frac{Y}{X-1})$

则

$\frac{a_0}{d}\ |\ \frac{\frac{Y}{X-1}}{d}\cdot (X^k-1)$

很明显$\frac{a_0}{d}和$$\frac{\frac{Y}{X-1}}{d}$是互质的

那么有

$\frac{a_0}{d}\ |\ (X^k-1)$

写成同余式的形式为

$X^k \equiv 1\ (mod\ \frac{a_0}{d})$

那么需要用到欧拉定理和The Luckiest number POJ3696这个题一模一样了

code：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
ll X,Y,a0;
ll mul(ll a,ll b,ll mod){
    ll ans = 0;
    while(b){
        if(b & 1)
            ans = (ans + a) % mod;
        a = (a + a) % mod;
        b >>= 1;
    }
    return ans;
}
ll q_pow(ll a,ll b,ll mod){
    ll ans = 1;
    while(b){
        if(b & 1)
            ans = mul(ans,a,mod);
        a = mul(a,a,mod);
        b >>= 1;
    }
    return ans;
}
ll Euler(ll n){
    ll ans = n;
    for(ll i = 2; i * i <= n; i++){
        if(n % i == 0){
            ans = ans / i * (i - 1);
            while(n % i == 0) n /= i;
        }
    }
    if(n > 1) ans = ans / n * (n - 1);
    return ans;
}
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
int main(){
    while(~scanf("%lld%lld%lld",&X,&Y,&a0)){
        ll p = a0 / gcd(a0,Y/(X-1));
        ll d = gcd(X,p);
        if(d != 1){
            printf("Impossible!\n");
        }
        else{
            ll phi = Euler(p);
            ll ans = phi;
            ll m = sqrt((double)phi);
            bool flag = false;
            for(ll i = 1; i <= m; i++){
                if(phi % i == 0 && q_pow(X,i,p) == 1){
                    ans = i;
                    flag = true;
                    break;
                }
            }
            if(!flag){
                for(ll i = m; i >= 2; i--){
                    if(phi % i == 0 && q_pow(X,phi/i,p) == 1){
                        ans = phi / i;
                        break;
                    }
                }
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}