HDU3307 Description has only two Sentences 欧拉函数 欧拉定理

 

 

Description has only two Sentences

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

 

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构造等比数列，先设p=y/(x-1);那么有(An+p)=x(An-1+p)  即 An+p=x^n(a0+p)

若An≡0（mod a0) 即 a0*x^n+p*(x^n - 1) ≡ 0 ≡ p*(x^n - 1)(mod a0)

同余方程两边同除gcds=gcd(p,a0),可得 p/gcds * (x^n - 1) ≡ 0 (mod a0/gcds)  *

∵ (p/gcds,a0/gcds) = 1

∴ *式可化为 x^n-1≡0 (mod ap=a0/gcds) 即 x^n ≡ 1 (mod ap)

根据欧拉定理可知 n = phi(ap) ，下面同反证法证明所求的k满足  k|n,k<=n

假设求得最小的满足题意的k 且 k \ n （记\为不整除）　　　

那么一定存在t使得k*t恰大于n，那么k*t-n也一定满足题意，但kt-n<k与条件矛盾，即k|n，k<=n

#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long lld;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
lld gcd(lld a,lld b){
    if(!b) return a;
    else   return gcd(b,a%b);
}

lld Min(lld a,lld b){return (a<b)?(a):(b);}

lld phi(lld n){
    lld ans=n;lld lim=sqrt(n);
    Rep(i,2,lim){
        if(n%i==0)  {
            ans=ans*(i-1)/i;
            while(n%i==0) n/=i;
        }
    }
    if(n>1) ans=ans*(n-1)/n;
    return ans;
}

lld x,y,a0,ap;

lld qpow(lld a,lld b){
    lld ans=1;
    while(b){
        if(b&1) ans=(ans*a)%ap;
        a=a*a%ap;
        b>>=1;
    }
    return ans%ap;
}

int main(){
    while(~scanf("%I64d%I64d%I64d",&x,&y,&a0)){
        lld p=y/(x-1);
        lld gcds=gcd(a0,p);
        ap=a0/gcds;
        if(gcd(x,ap)!=1){
            puts("Impossible!");
            continue;
        }
        lld ans=phi(ap);lld Lim=ans;
        Rep(i,2,sqrt(Lim))
          if(Lim%i==0){
              if(qpow(x,i)==1)  ans=Min(ans,i);
              if(qpow(x,Lim/i)==1) ans=Min(ans,Lim/i);
          }
        printf("%I64d\n",ans);
    }
    return 0;
}