POJ 3274 Gold Balanced Lineup

解题思路：hash

统计到当前位置为止，各位上出现1的个数，相对于最小次数的差值

比较出现相同差值序列的最长距离即可

欢迎指教

   
     

    
      
    
      #include
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      #define
    
       MAXN 100001
    
      

    
      #define
    
       PRIME 50101
    
      

    
      int
    
       feature[MAXN][
    
      30
    
      ],next[MAXN],first[PRIME];

    
      int
    
       main()
{
 
    
      int
    
       i,j,N,K,F,p,_min, _max
    
      =
    
      0
    
      ;
 
    
      long
    
       
    
      long
    
       t,t2;
 
    
      int
    
       bit[
    
      30
    
      ];
 
    
      for
    
      (i
    
      =
    
      0
    
      ,j
    
      =
    
      1
    
      ;i
    
      <
    
      30
    
      ;i
    
      ++
    
      ,j
    
      *=
    
      2
    
      )bit[i]
    
      =
    
      j;
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      MAXN;i
    
      ++
    
      )next[i]
    
      =-
    
      1
    
      ;
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      PRIME;i
    
      ++
    
      )first[i]
    
      =-
    
      1
    
      ;
 scanf(
    
      "
    
      %d %d
    
      "
    
      , 
    
      &
    
      N, 
    
      &
    
      K);
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      N;i
    
      ++
    
      )
 {
 scanf(
    
      "
    
      %d
    
      "
    
      ,
    
      &
    
      F);
 
    
      for
    
      (_min
    
      =
    
      MAXN,j
    
      =
    
      0
    
      ;j
    
      <
    
      K;j
    
      ++
    
      )
 {
 
    
      if
    
      (i
    
      >
    
      0
    
      )feature[i][j]
    
      =
    
      feature[i
    
      -
    
      1
    
      ][j];
 
    
      if
    
      (F
    
      &
    
      bit[j])feature[i][j]
    
      ++
    
      ;
 
    
      if
    
      (_min 
    
      >
    
       feature[i][j])_min
    
      =
    
      feature[i][j];
 }
 
    
      for
    
      (t
    
      =
    
      j
    
      =
    
      0
    
      ;j
    
      <
    
      K;j
    
      ++
    
      ){feature[i][j]
    
      -=
    
      _min;t2
    
      =
    
      feature[i][j];t
    
      +=
    
      (t2
    
      *
    
      t2);}
 
    
      if
    
      (t
    
      ==
    
      0
    
      &&
    
      _max
    
      <
    
      (i
    
      +
    
      1
    
      ))_max
    
      =
    
      i
    
      +
    
      1
    
      ;
 t
    
      %=
    
      PRIME;
 
    
      if
    
      (
    
      -
    
      1
    
      ==
    
      (p
    
      =
    
      first[t])){first[t]
    
      =
    
      i;
    
      continue
    
      ;}
 
    
      while
    
      (
    
      1
    
      )
 {
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <
    
      K;j
    
      ++
    
      )
    
      if
    
      (feature[p][j]
    
      !=
    
      feature[i][j])
    
      break
    
      ;
 
    
      if
    
      (j
    
      ==
    
      K
    
      &&
    
      ((i
    
      -
    
      p)
    
      >
    
      _max))_max
    
      =
    
      i
    
      -
    
      p;
 
    
      if
    
      (next[p]
    
      ==-
    
      1
    
      ){next[p]
    
      =
    
      i;
    
      break
    
      ;}
 p
    
      =
    
      next[p];
 }
 }
 printf(
    
      "
    
      %d\n
    
      "
    
      ,_max);
 
    
      return
    
       
    
      0
    
      ;
}