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=;

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

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

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

 begin
   readln(n);
   ans:=;
   for i:= 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.