poj 3378 Crazy Thairs

题意/Description：

    These days, Sempr is crazed on one problem named Crazy Thair. Given N (1 ≤ N ≤ 50000) numbers, which  are no more than 10⁹, Crazy Thair is a group of 5 numbers {i, j, k, l, m} satisfying:

    1. 1 ≤ i < j < k < l < m ≤ N

    2. A_i < A_j < A_k < A_l < A_m

  For example, in the sequence {2, 1, 3, 4, 5, 7, 6},there are four Crazy Thair groups: {1, 3, 4, 5, 6}, {2, 3, 4, 5, 6}, {1, 3, 4, 5, 7} and {2, 3, 4, 5, 7}.

Could you help Sempr to count how many Crazy Thairs in the sequence?

 

读入/Input：

  Input contains several test cases. Each test case begins with a line containing a number N, followed by a line containing N numbers.

 

输出/Output：

  Output the amount of Crazy Thairs in each sequence.

 

题解/solution：

  网上的解题报告有两个解法：

    1：Dp+线段树+离散化+高精度

    2：树状数组+Dp+离散化+高精度

  靠，好复杂，又没有P语言，C语言不会翻，I give up it。然后LZH经过了N天N夜，在 big head 的叽叽咕咕下，在 pig 的乱搞下，AC啦。于是我狠狠的敲了一波标。

  讲一下DP：

    F[I,j]表示用a[i]结尾的长度为j的序列数目。

    F[I,j]= sum(F[k,j-1]) (1<=k<I且a[i]>a[k])

  因为读入的数有10^9大，而数列长度只有50000，想到离散化。可离散后，位置发生改变，排个序，在二分来查找。

要找它前方的所有比它小的树的t-1之和，想到了单点更新区间查询，线段树和数状数组都可以维护。由于长度为5，用5个树。完成这些，会发现结果会爆int64，于是加个高精度。看完后，你肯定。

 

代码/Code：

type
  arr=record
    x,y:longint;
  end;

var
  dp:array [0..50001,1..5] of int64;
  m,len,n,k:longint;
  f:array [0..50001] of longint;
  tree:array [0..50001] of arr;
  sum:array [0..101] of longint;

procedure qsort(l,r:longint);
var
  i,j,key,key1:longint;
  temp:arr;
begin
  if l>=r then exit;
  i:=l; j:=r;
  key:=tree[(l+r) shr 1].x;
  key1:=tree[(l+r) shr 1].y;
  repeat
    while (tree[i].x<key) or (tree[i].x=key) and (tree[i].y<key1) do inc(i);
    while (tree[j].x>key) or (tree[j].x=key) and (tree[j].y>key1) do dec(j);
    if i<=j then
      begin
        temp:=tree[i]; tree[i]:=tree[j]; tree[j]:=temp;
        inc(i);dec(j);
      end;
  until i>j;
  qsort(l,j);
  qsort(i,r);
end;

function bit(n:longint):longint;
begin
    exit(n and -n);
end;

procedure jf(n:qword);
var
  i:longint;
  a,b:array [0..100] of longint;
begin
  fillchar(a,sizeof(a),0);
  fillchar(b,sizeof(b),0);
  i:=-1;
  while n>0 do
    begin
      inc(i);
      a[i]:=n mod 10;
      n:=n div 10;
    end;
  i:=-1;
  while i<100 do
    begin
      inc(i);
      b[i]:=a[i]+sum[i]+b[i];
      if b[i]>=10 then
        begin
          inc(b[i+1]);
          b[i]:=b[i] mod 10;
        end;
    end;
  for i:=0 to 99 do
    sum[i]:=b[i];
end;

function count(n,j:longint):int64;
var
  ans:int64;
begin
  ans:=0;
  while n>0 do
    begin
      ans:=ans+dp[n,j];
      n:=n-bit(n);
    end;
  exit(ans);
end;

procedure update(n,j:longint;k:int64);
begin
  while n<=m do
    begin
      dp[n,j]:=dp[n,j]+k;
      n:=n+bit(n);
    end;
end;

procedure dpp(n:longint);
var
  tem:int64;
  i,j:longint;
begin
  fillchar(dp,sizeof(dp),0);
  fillchar(sum,sizeof(sum),0);
  len:=1;
  for i:=1 to n do
    begin
      tem:=count(f[i]-1,4);
      jf(tem);
      for j:=5 downto 2 do
        begin
          tem:=count(f[i]-1,j-1);
          update(f[i],j,tem);
        end;
      update(f[i],1,1);
    end;
  len:=100;
  while(sum[len]=0) and (len>0) do dec(len);
  for i:=len downto 0 do
    write(sum[i]);
  writeln;
end;

procedure main;
var
  i:longint;
begin
  while not eof do
    begin
      readln(n);
      m:=n;
      for i:=1 to n-1 do
        begin
          read(tree[i].x);
          tree[i].y:=i;
        end;
      readln(tree[n].x);
      tree[n].y:=n;
      qsort(1,n);
      f[tree[1].y]:=1; k:=0;
      for i:=1 to n do
        begin
          if tree[i].x=tree[i-1].x then f[tree[i].y]:=f[tree[i-1].y] else
            begin
              f[tree[i].y]:=k+1;
              inc(k);
            end;
        end;
      dpp(n);
    end;
end;

begin
  main;
end.