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