BZOJ3724 [HNOI2012]集合选数 【状压dp】

题目链接

BZOJ3724

题解

构造矩阵的思路真的没想到
选\(x\)就不能选\(2x\)和\(3x\)，会发现实际可以转化为矩阵相邻两项
\[\begin{matrix}1 & 3 & 9 & 27 & ... \\2 & 6 & 18 & 54 & ... \\4 & 12 & 36 & 108 & ... \\8 & 24 & 72 & 216 & ... \\ ... & ... &... &... &... \\ \end{matrix}\]
相当于选这样的矩阵中不相邻的若干项的方案数
我们取每一个不是\(2\)和\(3\)的倍数的数作为矩阵左上角
行数和列数都很小，可以状压\(dp\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 20,maxm = (1 << 12),INF = 1000000000,P = 1000000001;
int f[maxn][maxm];
int N,n,cnt[maxn];
int ill[maxm],ans = 1;
void init(){
    int t = 3;
    for (int s = 0; s < maxm; s++){
        for (int i = 0; i <= 10; i++)
            if (((s & (t << i)) >> i) == t){
                ill[s] = true; break;
            }
    }
}
void dp(int x){
    for (n = 0; x <= N; x *= 2){
        cnt[++n] = 0;
        int t = x;
        while (t <= N) cnt[n]++,t *= 3;
    }
    for (int i = 1; i <= n; i++)
        for (int s = 0; s < (1 << cnt[i]); s++)
            f[i][s] = 0;
    for (int s = 0; s < (1 << cnt[1]); s++)
        if (!ill[s]) f[1][s] = 1;
    for (int i = 2; i <= n; i++)
        for (int s = 0; s < (1 << cnt[i]); s++){
            if (ill[s]) continue;
            for (int e = 0; e < (1 << cnt[i - 1]); e++){
                if (ill[e]) continue;
                if (!(s & e)){
                    f[i][s] = (f[i][s] + f[i - 1][e]) % P;
                }
            }
        }
    int re = 0;
    for (int s = 0; s < (1 << cnt[n]); s++)
        re = (re + f[n][s]) % P;
    ans = 1ll * ans * re % P;
}
int main(){
    init();
    scanf("%d",&N);
    for (int i = 1; i <= N; i++){
        if (i % 2 == 0 || i % 3 == 0) continue;
        dp(i);
    }
    printf("%d\n",ans);
    return 0;
}