"Pressure DP" bzoj2734-[hnoi2012) set of selection

Already August the pill, start liver work and prepare the college entrance examination!!

"The main topic"

"Set theory and Graph theory" This course has a homework problem, ask the students to find {1, 2, 3, 4, 5} All meet the following conditions of the subset: If X is in this subset, 2x and 3x cannot be in the sub-set. Now ask the following question: For any positive integer n≤100000, how to find the number of subsets of {1, 2,..., n} that meet the above constraints (just output the result of 1,000,000,001 modulo) (including the empty set).

Ideas

For an integer x of any and 6 coprime within N, we list A matrix:

X 3x 9x 27x ...

2x 6x 18x 54x ...

4x 12x 36x 108x ...

So we now enumerate this number x with 6 coprime and then proceed to the transition of State compression. This is somewhat similar to the previous king. F[I][J] represents the total possibility of line I, and the status of Line I is J. However, it is not necessarily a rectangle, the number of columns per row may be different, and for a number of rows J, we only need to enumerate 0. The status of the 2^j-1 and is recorded as before transferred to the next line of DP.

Here's a scrolling array, but don't forget to empty each new scrolling array.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 #defineMoD 10000000016 using namespacestd;7typedefLong LongLL;8 Const intmaxn= A;9 intN;Ten intusable[1<<maxn],f[2][1<<MAXN]; One BOOLmark[1<<MAXN]; A  - intUsable (intx) - { the     if(x<<1&x | | X>>1&AMP;X)return 0;Else return 1; - } -  - intdpintNow ) + { -Memset (F,0,sizeof(f)); +     intCur=0, before=-1;//before refers to the previous line with several numbers A      for(intI=0; now* (1<<i) <=n;i++)//enumerate the first number of each row to find the total number of rows at     { -Cur=1-cur; -         inttmp=now* (1<<i), J; -          for(j=0; tmp<=n;j++,tmp*=3);//find out how many numbers each line has -          for(intk=0;k< (1&LT;&LT;J); k++)//enumerates the status of the current row -         { inf[cur][k]=0;//"Don't forget to initialize ☆" -             if(Usable[k]) to             { +                 if(before==-1) {f[cur][k]=1;Continue;}//If this is the first row, set the feasible status to 1 -                  for(intp=0;p < (1<<before);p + +) the                     if(Usable[p]) *                         if((k&p) = =0) f[cur][k]=f[cur][k]+f[1-cur][p],f[cur][k]%=mod;//don't forget about the mod here. $             }Panax Notoginseng         } -Before=J; the     } +     intans=0; A      for(intI=0;i< (1<<before); i++) ans+=f[cur][i],ans%=MoD; the     return(ans); + } -  $ voidgetusable () $ { -memset (Usable,0,sizeof(usable)); -      for(intI=0;i< (2&LT;&LT;MAXN); i++) the         if(Usable (i)) usable[i]=1; - }Wuyi  the voidSolve () - { Wumemset (Mark,0,sizeof(Mark)); -LL ans=1;//in order to prevent the multiplication of overflow, you can first use Longlong, and then converted back to int About      for(intI=1; i<=n;i++) $         if((i%2) && (i%3)) ans= (ANS*DP (i))%MoD; -printf"%d\n",(int) ans); - } -  A intMain () + { thescanf"%d",&n); - getusable (); $ solve (); the     return 0; the}

"Pressure DP" bzoj2734-[hnoi2012) set of selection