bzoj2734 "HNOI2012" collection selection

2734: [HNOI2012] Collection selection time limit: ten Sec Memory Limit: MB
Submit: 831 Solved: 487
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"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. The students do not like the problem with the enumeration nature, so it becomes the following question: For any positive integer n≤100000, how to find out {1, 2,..., n} The number of the subset of the constraints to meet the above constraint (just output the results of 1,000,000,001 modulo), Now this problem is yours.

Input

Only one row, where there is a positive integer n,30% the data satisfies n≤20.

Output

Contains only a positive integer representing the number of subsets of {1, 2,..., n} that meet the above constraints.

Sample Input
4Sample Output8

"Sample Interpretation"

There are 8 sets that meet the requirements, namely the empty set, {1},{1,4},{2},{2,3},{3},{3,4},{4}.
HINT

Source

Day2

pressure DP idea good problem

Write such a matrix

1 3 9 27 ...

2 6 18 54 ...

4 12 36 108 ...

...

You can find a maximum of 12 rows.

So we just enumerate the upper left corner of the number x, we can get a different matrix, for each matrix need to select some number, but not the number of adjacent, the pressure DP solution.

For each matrix, multiplying the scheme number is the answer.

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib > #include <algorithm> #define F (I,j,n) for (int. i=j;i<=n;i++) #define D (i,j,n) for (int i=j;i>=n;i--) # Define ll long long#define MAXN 100005#define mod 1000000001using namespace std;int n;ll Ans=1,f[20][4100],num[20];bool vs T[maxn];inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} return x*f;} inline ll Calc (int x) {int cnt=0;memset (num,0,sizeof (num)); memset (F,0,sizeof (f)); F[0][0]=1;while (x<=n) {Cnt++;int Tmp=x;while (tmp<=n) {num[cnt]++;vst[tmp]=true;tmp*=3;} F (i,0, (1<<num[cnt])-1) {int p;for (p=1;p<num[cnt];p + +) if ((i& (1<<)) && (P-1 (i&) <p)) break;if (p!=num[cnt]) continue; F (j,0, (1<<num[cnt-1])-1) if (! ( I&AMP;J)) (F[cnt][i]+=f[cnt-1][j])%=mod;} x*=2;} ll t=0; F (i,0, (1<<num[cnt])-1) (T+=f[cnt][i])%=mod;return T;} int main () {memset (vst,false,sizeof (VST)); N=read (); F (I,1,n) if (!vst[i]) (Ans*=calc (i))%=mod;printf ("%lld\n", ans); return 0;}

bzoj2734 "HNOI2012" collection selection