题目链接
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
[egin{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
",ans);
return 0;
}