【Codeforces Round #228 (Div.1）】—D. Fox and Perfect Sets（线性基+数位dp）

传送门

考虑实际上就是构造一个线性基使得最大数不超过$k$
令$f[i][j][0/1]$表示前$i$位，已经有$j$个基，是否抵着上界

考虑当前位是否加入一个新的基
如果不加，则考虑前面$j$个异或起来这一位是否为$1$
为$1,0$的方案都是$2^{j-1}$

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),a);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=36;
int k,p[N],len,f[N][N][2];
int main(){
	k=read();
	while(k){
		p[++len]=k%2;
		k/=2;
	}
	reverse(p+1,p+len+1);
	f[0][0][1]=1;
	for(int i=1;i<=len;i++)
	for(int j=0;j<=len;j++){
		int x=j?(1<<(j-1)):1,y=j?(1<<(j-1)):0;
		Add(f[i][j][0],mul(f[i-1][j][0],1<<j));
		Add(f[i][j+1][0],f[i-1][j][0]);
		if(p[i]){
			Add(f[i][j][0],mul(f[i-1][j][1],x));
			Add(f[i][j][1],mul(f[i-1][j][1],y));
			Add(f[i][j+1][1],f[i-1][j][1]);
		}
		else Add(f[i][j][1],mul(f[i-1][j][1],x));
	}
	int res=0;
	for(int i=0;i<=len;i++)Add(res,add(f[len][i][0],f[len][i][1]));
	cout<<res;
}