hdu 3307 Description has only two Sentences (欧拉函数+快速幂)

Description has only two Sentences
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 852 Accepted Submission(s): 259


Problem Description
an = 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.


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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因为考试放下了挺久，后来发现不做题好空虚寂寞...于是决定在做一段时间再说。

//31MS    236K    1482 B    G++
/*

    又是不太懂的数学题，求ak，令ak%a0==0
     
    欧拉函数+快速幂：
         欧拉函数相信都知道了。
         首先这题推出来的公式为：
             ak=a0+y/(x-1)*(x^k-1);
         
         明显a0是可以忽略的，其实就是要令 
             y/(x-1)*(x^k-1) % a0 == 0;
         可令 m=a0/(gcd(y/(x-1),a0))，然后就求k使得
             (x^k-1)%m==0 即可
             即 x^k==1(mod m)
         
         又欧拉定理有：
              x^euler(m)==1(mod m)  (x与m互质，不互质即无解)
         
         由抽屉原理可知 x^k 的余数必在 euler(m) 的某个循环节循环。
         故求出最小的因子k使得 x^k%m==1，即为答案 

*/
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
__int64 e[1005],id;
int cmp(const void*a,const void*b)
{
    return *(int*)a-*(int*)b;
}
__int64 euler(__int64 n)
{
    __int64 m=(__int64)sqrt(n+0.5);
    __int64 ret=1;
    for(__int64 i=2;i<=m;i++){
        if(n%i==0){
            ret*=i-1;n/=i; 
        }
        while(n%i==0){
            ret*=i;n/=i;
        }
    }
    if(n>1) ret*=n-1;
    return ret;
}
__int64 Gcd(__int64 a,__int64 b)
{
    return b==0?a:Gcd(b,a%b);
}
void find(__int64 n)
{
    __int64 m=(__int64)sqrt(n+0.5);
    id=0;
    for(__int64 i=1;i<m;i++)
        if(n%i==0){
            e[id++]=i;
            e[id++]=n/i;
        }
    if(m*m==n) e[id++]=m;
}
__int64 Pow(__int64 a,__int64 b,__int64 mod)
{
    __int64 t=1;
    while(b){
        if(b&1) t=(t*a)%mod;
        a=(a*a)%mod;
        b/=2;
    } 
    return t;
} 
int main(void)
{
    __int64 x,y,a;
    while(scanf("%I64d%I64d%I64d",&x,&y,&a)!=EOF)
    {
        if(y==0){
            puts("1");
            continue;
        } 
        __int64 m=a/(Gcd(y/(x-1),a));
        if(Gcd(m,x)!=1){
            puts("Impossible!");
            continue;
        }
        __int64 p=euler(m);
        find(p);
        qsort(e,id,sizeof(e[0]),cmp);
        for(int i=0;i<id;i++){
            if(Pow(x,e[i],m)==1){
                printf("%I64d
",e[i]);
                break;
            }
        }
    }
    return 0;    
}