【题解】codeforces 1065G. Fibonacci Suffix 合并技巧+按位贪心

题目连接

题意：

给出一个斐波那契序列，F[0] = “0” ， F[1] = “1” ， F[i] = F[i - 2] + F[i - 1]. 求第k小的后缀的前m位。

题解

直接按位贪心
每次check一个前缀在F[n]中的出现次数。
这个可以维护pre，suf , num表示F[i]和当前串的前后缀匹配长度和当前串出现次数
长度不够默认可以匹配，用bitset优化合并
复杂度O(n * m * m/32)
注意特判当前串是否作为一个后缀出现，因为总是把后面的串作为后缀，暴力跑200位然后判断即可
还有几个坑点：
出现次数可能很大，会爆longlong，要和inf取min
注意每次check的清空和初始化

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define forup(i,l,r) for (register int i = l ; i <= r ; i += lowbit(i))
#define fordown(i,id) for (register int i = id ; i ; i -= lowbit(i))
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define lowbit(x) (x&(-x))

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int,int> pr;

const ll inf = 1e18;
const int N = 220;
const int maxn = 220;
const ll mod = 1e9 + 7;

struct node{
	ll num,len;
	bitset <N> pre,suf;
	node() { num = len = 0; pre.reset(),suf.reset(); }
	void clear(){
		num = len = 0;
		pre.reset() , suf.reset();
	}
}f[220];
int fail[maxn];
struct node2{
	int len;
	int s[maxn * 10];
	node2(){
		len = 0;
	}
	void init(int x){
		s[++len] = x;
	}
	node2 operator + (const node2 &a){
		node2 res;
		rep(i,1,len) res.s[i] = s[i];
		rep(i,1,a.len) res.s[i + len] = a.s[i];
		res.len = a.len + len;
		return res;
	}
	void getfail(int m,int a[]){
		fail[1] = 0;
		int p = 0;
		rep(i,2,m){
			while ( p && a[p + 1] != a[i] ) p = fail[p];
			if ( a[p + 1] == a[i] ) p++;
			fail[i] = p;
		}
	}
	bool check(int m,int a[]){
	//	getfail(m,a);
	//	int p = 0;
	//	rep(i,1,len){
	//		while ( p && s[i] != a[p + 1] ) p = fail[p];
	//		if ( a[p + 1] == a[i] ) p++;
	//		if ( p == m ) return 1;
	//	}
	//	return 0;
		rep(i,1,m) if ( a[i] != s[len - m + i] ) return 0;
		return 1;
	}
}dt[maxn];

bitset <N> all;
int n,m,ans[maxn],id;
ll k;
int len;

node merge(const node &a,const node &b){
	if ( !a.len ) return b;
	node res;
	res.len = a.len + b.len;
	res.pre = a.pre , res.suf = b.suf; //pre和suf中包含了长度1..n-1的所有信息，所以合并的时候直接&就好
	res.num = a.num + b.num;
	if ( a.len <= len - 2 ) res.pre &= (b.pre >> a.len) | (all << (len - 1 - a.len)); 
	if ( b.len <= len - 2 ) res.suf &= (a.suf << b.len) | (all >> (len - 1 - b.len));
	if ( res.len >= len ){
		bitset<N> tmp = a.suf & b.pre;
		if ( a.len <= len - 2 ) tmp &= all >> (len - 1 - a.len); //如果a的长度不足，则删除高位
		if ( b.len <= len - 2 ) tmp &= all << (len - 1 - b.len); //删除低位
		res.num += tmp.count();
	}
	if ( res.num > inf ) res.num = inf;
	return res;
}
void init(){
	dt[0].init(0) , dt[1].init(1);
	rep(i,2,100){
		dt[i] = dt[i - 2] + dt[i - 1];
		if ( dt[i].len > m ){ id = i; break; }
	}
//	rep(i,1,dt[id].len) printf("%d",dt[id].s[i]);
//	puts("");
}

inline ll calc(int a[]){
	all.reset();
	rep(i,1,len - 1) all.set(i);
	rep(i,0,n) f[i].clear();
	f[0].len = 1 , f[1].len = 1;
	if ( len == 1 ){
		f[0].num = a[1] == 0;
		f[1].num = a[1] == 1;
	}
	else{
		rep(i,1,len){
			f[0].pre[i - 1] = a[i] == 0;
			f[0].suf[i] = a[i] == 0;
			f[1].pre[i - 1] = a[i] == 1;
			f[1].suf[i] = a[i] == 1;
		}
	}
	rep(i,2,n) f[i] = merge(f[i - 2] ,f[i - 1]);
	return f[n].num;
}
void solve(int n,ll k){
	if ( n > m ){ len = n - 1; return; }
	ans[len = n] = 0;
	ll num = calc(ans);
//	cout<<num<<endl;
	if ( num >= k ){
		if ( dt[id].check(n,ans) ) k--;
		if ( !k ){ len = n; return; }
		solve(n + 1,k);
	}
	else{
		k -= num;
		ans[n] = 1;
		if ( dt[id].check(n,ans) ) k--;
		if ( !k ){ len = n; return; }
		solve(n + 1,k);
	}
}
int main(){
	scanf("%d %lld %d",&n,&k,&m);
	init();
	solve(1,k);
	rep(i,1,len) printf("%d",ans[i]);
   puts("");	
}