Problem--HOJ1372 方程的解数(Hash)

Problem--HOJ1372 方程的解数(Hash)

Analysis:This problem requires that we find the number of solutions to an equation.The equation is given as k ₁*x ₁^q ₁+k ₂*x ₂^q ₂+...+k _n*x _n^q _n=0,where n is within the range [1,6],x _i is within the range [1,150],and q _i is guaranteed to be positive integers.Besides,the number we're dealing with will not exceed the range of integer.
Naturally,we will try to enumerate all the possible values for x _i and see if each of them satisfy the equation.However,this will cause us quite a lot of time.With a little improvement,we can move half of the items of the polynomial to the other side of the equation,and form a new one like k ₁*x ₁^q ₁+k ₂*x ₂^q ₂+...k _i*x _i^q _i=-(k _i+1*x _i+1^q _i+1+k _i+2*x _i+2^q _i+2+...k _n*x _n^q _n).With this equation at hand,all we need to do is to build a table of value of the left side and try to find out how many combinations of x _i on the other side can make a value existing in the table we have.Consequently,we need the Hash table.The strategy we use here to deal with the collision would be to build a list on each node of the table.More details in the code.
Code:

  1  #include < cstdio >
  2  #include < cstring >
  3  #include < cmath >
  4  #include < vector >
  5  #define  MAXSIZE 500050
  6  #define  MOD 400037
  7  using   namespace  std;
  8  struct  answer
  9  {
 10       int  value,num;
 11      answer *  next;
 12  }pool[MAXSIZE * 4 ];
 13  answer  * htable[MAXSIZE], * pp, * pt;
 14  answer *  create( int  value, int  num,answer *  temp)
 15  {
 16      pp -> value = value;pp -> num = num;
 17      pp -> next = temp; return  pp ++ ;
 18  }
 19   
 20  int  powers[ 151 ][ 6 ];
 21  int  t[ 6 ],coe[ 6 ];
 22  int  n,m,sum,counter;
 23  void  init( int  n, int  m)
 24  {
 25       for ( int  i = 1 ; i <= m; i ++ )
 26      {
 27           for ( int  j = 0 ; j < n; j ++ )
 28          {
 29              powers[i][j] = ( int )pow(( double )i,t[j]);
 30          }
 31      }
 32  }
 33  void  init_hash()
 34  {
 35      memset(htable, 0 , sizeof (htable));
 36      pp = pool;
 37  }
 38  void  insert_hash( int  sum)
 39  {
 40       int  t = (sum > 0 )  ?  sum : ( 0 - sum);t %= MOD;
 41       for (pt = htable[t];pt;pt = pt -> next)
 42      {
 43           if (pt -> value == sum)
 44          {
 45              pt -> num ++ ;
 46               return ;
 47          }
 48      }
 49      htable[t] = create(sum, 1 ,htable[t]);
 50  }
 51  void  get_hash( int  sum)
 52  {
 53       int  t = (sum > 0 )  ?  sum : ( 0 - sum);t %= MOD;
 54       for (pt = htable[t];pt;pt = pt -> next)
 55      {
 56           if (pt -> value == sum)
 57          {
 58              counter += pt -> num;
 59               return ;
 60          }
 61      }
 62  }
 63  void  dfs( int  depth, int  m)
 64  {
 65       if (depth == 0 )
 66      {
 67          insert_hash(sum);
 68           return ;
 69      }
 70       for ( int  i = 1 ; i <= m; i ++ )
 71      {
 72           int  tp = (coe[depth - 1 ] * powers[i][depth - 1 ]);
 73          sum += tp;
 74          dfs(depth - 1 ,m);
 75          sum -= tp;
 76      }
 77  }
 78  void  get_result( int  d, int  m)
 79  {
 80       if ( ! d)
 81      {
 82          get_hash( 0 - sum);
 83           return ;
 84      }
 85       for ( int  i = 1 ; i <= m; i ++ )
 86      {
 87           int  tp = (coe[n - d] * powers[i][n - d]);
 88          sum += tp;
 89          get_result(d - 1 ,m);
 90          sum -= tp;
 91      }
 92  }
 93  int  main()
 94  {
 95       int  dp;
 96       while (scanf( " %d " , & n) == 1 )
 97      {
 98          scanf( " %d " , & m);
 99           for ( int  i = 0 ; i < n; i ++ ) scanf( " %d%d " , & coe[i], & t[i]);
100           if (n == 1 ) printf( " 0\n " );
101           else
102          {
103              init(n,m);
104              init_hash();
105               if (n == 2 ) dp = 1 ;
106               else   if (n < 6 ) dp = 2 ;
107               else  dp = 3 ;
108   
109              sum = 0 ;
110              dfs(dp,m);
111              sum = counter = 0 ;
112              get_result(n - dp,m);
113              printf( " %d\n " ,counter);
114          }
115      }
116  }