ACM_高次同余方程

来源：互联网发布：linux memset 编辑：程序博客网时间：2024/04/29 03:34
/*poj 3243  *解决高次同余方程的应用，已知 X^Y = K mod Z, 及X,Z,K的值，求 Y 的值 */#include<cstdio>  #include<cstring>  #include<cmath>  using namespace std;  #define lint __int64  #define MAXN 131071  struct HashNode { lint data, id, next; };  HashNode hash[MAXN<<1];  bool flag[MAXN<<1];  lint top;    void Insert ( lint a, lint b )  {      lint k = b & MAXN;      if ( flag[k] == false )      {          flag[k] = true;          hash[k].next = -1;          hash[k].id = a;          hash[k].data = b;          return;      }      while( hash[k].next != -1 )      {          if( hash[k].data == b ) return;          k = hash[k].next;      }      if ( hash[k].data == b ) return;      hash[k].next = ++top;      hash[top].next = -1;      hash[top].id = a;      hash[top].data = b;  }    lint Find ( lint b )  {      lint k = b & MAXN;      if( flag[k] == false ) return -1;      while ( k != -1 )      {          if( hash[k].data == b ) return hash[k].id;          k = hash[k].next;      }      return -1;  }    lint gcd ( lint a, lint b )  {      return b ? gcd ( b, a % b ) : a;  }    lint ext_gcd (lint a, lint b, lint& x, lint& y )  {      lint t, ret;      if ( b == 0 )      {          x = 1, y = 0;          return a;      }      ret = ext_gcd ( b, a % b, x, y );      t = x, x = y, y = t - a / b * y;      return ret;  }    lint mod_exp ( lint a, lint b, lint n )  {      lint ret = 1;      a = a % n;      while ( b >= 1 )      {          if( b & 1 )              ret = ret * a % n;          a = a * a % n;          b >>= 1;      }      return ret;  }    lint BabyStep_GiantStep ( lint A, lint B, lint C )  {      top = MAXN;  B %= C;      lint tmp = 1, i;      for ( i = 0; i <= 100; tmp = tmp * A % C, i++ )          if ( tmp == B % C ) return i;        lint D = 1, cnt = 0;      while( (tmp = gcd(A,C)) !=1 )      {          if( B % tmp ) return -1;          C /= tmp;          B /= tmp;          D = D * A / tmp % C;          cnt++;      }        lint M = (lint)ceil(sqrt(C+0.0));      for ( tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++ )          Insert ( i, tmp );        lint x, y, K = mod_exp( A, M, C );      for ( i = 0; i <= M; i++ )      {          ext_gcd ( D, C, x, y ); // D * X = 1 ( mod C )          tmp = ((B * x) % C + C) % C;          if( (y = Find(tmp)) != -1 )              return i * M + y + cnt;          D = D * K % C;      }      return -1;  }    int main()  {      lint A, B, C;      while( scanf("%I64d%I64d%I64d",&A,&C,&B ) !=EOF )      {          if ( !A && !B && !C ) break;          memset(flag,0,sizeof(flag));          lint tmp = BabyStep_GiantStep ( A, B, C );          if ( tmp == -1 )puts("No Solution");          else printf("%I64d\n",tmp);      }      return 0;  }
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