[NDK 出栈序列统计]

[题目来源]：NOIP基础题目集

[关键字]：数学

[题目大意]：求出n个数的出栈序列数目

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[分析]：n个数的出栈序列数目就是一颗有n个节点的二叉树的数量。因为他的进栈序列就是该二叉树的先序遍历，而它的出栈序列就是该二叉树的中序遍历，所以两者都等于Catalan数公式为：C（2n，n）/（n+1）。因为当n稍微大一点时结果就会很大，所以要用高精度，为了避免高精除法，可以先将把分子和分母分别写成几个质数相乘的形式，再约分（指数相减）。

[代码]：

View Code

  1 program Project1;
  2 type
  3   rec = record
  4     len: longint;
  5     dat: array[0..2000] of longint;
  6   end;
  7 var
  8   ans: rec;
  9   d: array[0..200] of longint;
 10   su: array[0..200] of longint;
 11   n, tot: longint;
 12 
 13 procedure cheng(var ans: rec; x: longint);
 14 var
 15   i, k: longint;
 16 begin
 17   k := 0;
 18   for i := 1 to ans.len do
 19     begin
 20       ans.dat[i] := ans.dat[i]*x+k;
 21       k := ans.dat[i] div 10;
 22       ans.dat[i] := ans.dat[i] mod 10;
 23     end;
 24   while k > 0 do
 25     begin
 26       inc(ans.len);
 27       ans.dat[ans.len] := k mod 10;
 28       k := k div 10;
 29     end;
 30 end;
 31 
 32 procedure work;
 33 var
 34   i, j, k: longint;
 35 begin
 36   for i := n+2 to 2*n do
 37     begin
 38       j := 0;
 39       k := i;
 40       while k > 1 do
 41         begin
 42           inc(j);
 43           while k mod su[j] = 0 do
 44             begin
 45               inc(d[j]);
 46               k := k div su[j];
 47             end;
 48         end;
 49     end;
 50   for i := 2 to n do
 51     begin
 52       j := 0;
 53       k := i;
 54       while k > 1 do
 55         begin
 56           inc(j);
 57           while k mod su[j] = 0 do
 58             begin
 59               dec(d[j]);
 60               k := k div su[j];
 61             end;
 62         end;
 63     end;
 64   ans.len := 1;
 65   ans.dat[1] := 1;
 66   for i := 1 to tot do
 67     while d[i] > 0 do
 68       begin
 69         cheng(ans,su[i]);
 70         dec(d[i]);
 71       end;
 72   for i := ans.len downto 1 do
 73     write(ans.dat[i]);
 74 end;
 75 
 76 procedure make;
 77 var
 78   i, j: longint;
 79   f: boolean;
 80 begin
 81   for i := 2 to 2*n do
 82     begin
 83       f := true;
 84       for j := 2 to i-1 do
 85         if i mod j = 0 then
 86           begin f := false; break; end;
 87        if f then
 88          begin inc(tot); su[tot] := i; end;
 89     end;
 90 end;
 91 
 92 begin
 93   assign(input,'d:\in.txt');reset(input);
 94   assign(output,'d:\out.txt');rewrite(output);
 95   read(n);
 96   make;
 97   work;
 98   close(input);
 99   close(output);
100 end.