bzoj2734

非常巧妙地题目
对于一个数x
列出这样的矩阵
x 2x 4x 8x ……
3x 6x 12x 24x ……
…………………………
不难方案数就是求取数不相邻的方案数
考虑矩阵宽不超过logn，所以可以用状压dp解决

const mo=1000000001;

var f:array[0..18,0..100010] of longint;
    s:array[0..18] of longint;
    v:array[0..100010] of boolean;
    i,n:longint;
    ans:int64;

function dp(x:longint):int64;
  var i,j,k,y:longint;
  begin
    dp:=0;
    y:=x;
    v[x]:=true;
    s[1]:=1;
    while y*3<=n do
    begin
      y:=y*3;
      v[y]:=true;
      inc(s[1]);
    end;
    for j:=0 to 1 shl s[1]-1 do
      if j and (j shl 1)=0 then f[1,j]:=1;

    i:=1;
    while x*2<=n do
    begin
      inc(i);
      x:=x*2;
      y:=x;
      s[i]:=1;
      v[y]:=true;
      while y*3<=n do
      begin
        y:=y*3;
        v[y]:=true;  //避免重复
        inc(s[i]);
      end;
      for j:=0 to 1 shl s[i]-1 do
        f[i,j]:=0;
      for j:=0 to 1 shl s[i]-1 do
        if j and (j shl 1)=0 then
          for k:=0 to 1 shl s[i-1]-1 do
            if j and k=0 then f[i,j]:=(f[i,j]+f[i-1,k]) mod mo;
    end;
    for j:=0 to 1 shl s[i]-1 do
      inc(dp,f[i,j]);
  end;

begin
  readln(n);
  ans:=1;
  for i:=1 to n do
    if not v[i] then
      ans:=ans*dp(i) mod mo; //这显然是乘法原理
  writeln(ans);
end.