题目链接

Solution 斐波那契

题目大意：兔子按照斐波那契提出的模型繁衍，根据它们的父子关系形成一棵树。每次给定两个点，求出它们的 \(lca\)。

贪心

分析：首先暴力建树肯定是会直接上天的

我们发现，对于任何一个不为根的节点，它和它的父亲编号的差值是一个斐波那契数。

证明可以参考 斐波那契进制，LeetCode 类似问题的题解（我是Graphviz找规律发现的）

因此，对于一个数，我们减去它在斐波那契数列里面的前驱，就可以得到它的父亲。

树高不超过 \(60\)，直接暴力求 \(lca\) 就可以了。

#include <cstdio>
#include <cctype>
#include <algorithm>
#include <vector>
#pragma GCC optmize(2)
using namespace std;
typedef long long ll;
struct IO{//-std=c++11,with cstdio and cctype
	private:
		static constexpr int ibufsiz = 1 << 20;
		char ibuf[ibufsiz + 1],*inow = ibuf,*ied = ibuf;
		static constexpr int obufsiz = 1 << 20;
		char obuf[obufsiz + 1],*onow = obuf;
		const char *oed = obuf + obufsiz;
	public:
		inline char getchar(){
			#ifndef ONLINE_JUDGE
				return ::getchar();
			#else
				if(inow == ied){
					ied = ibuf + sizeof(char) * fread(ibuf,sizeof(char),ibufsiz,stdin);
					*ied = '\0';
					inow = ibuf;
				}
				return *inow++;
			#endif
		}
		template<typename T>
		inline void read(T &x){
			static bool flg;flg = 0;
			x = 0;char c = getchar();
			while(!isdigit(c))flg = c == '-' ? 1 : flg,c = getchar();
			while(isdigit(c))x = x * 10 + c - '0',c = getchar();
			if(flg)x = -x;
		}
		template <typename T,typename ...Y>
		inline void read(T &x,Y&... X){read(x);read(X...);}
		inline int readi(){static int res;read(res);return res;}
		inline long long readll(){static long long res;read(res);return res;}
		
		inline void flush(){
			fwrite(obuf,sizeof(char),onow - obuf,stdout);
			fflush(stdout);
			onow = obuf;
		}
		inline void putchar(char c){
			#ifndef ONLINE_JUDGE
				::putchar(c);
			#else
				*onow++ = c;
				if(onow == oed){
					fwrite(obuf,sizeof(char),obufsiz,stdout);
					onow = obuf;
				}
			#endif
		}
		template <typename T>
		inline void write(T x,char split = '\0'){
			static unsigned char buf[64];
			if(x < 0)putchar('-'),x = -x;
			int p = 0;
			do{
				buf[++p] = x % 10;
				x /= 10;
			}while(x);
			for(int i = p;i >= 1;i--)putchar(buf[i] + '0');
			if(split != '\0')putchar(split);
		}
		inline void lf(){putchar('\n');}
		~IO(){
			fwrite(obuf,sizeof(char),onow - obuf,stdout);
		}
}io;

constexpr ll fib[] = {1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025,20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041,1548008755920};

vector<ll> veca,vecb,tmp;
inline void resolve(vector<ll> &vec,ll x){
	tmp.clear();vec.clear();
	x--;
	while(x){
		const ll now = *prev(upper_bound(begin(fib),end(fib),x));
		tmp.push_back(now);
		x -= now; 
	}
	reverse(tmp.begin(),tmp.end());
	ll now = 1;
	vec.push_back(now);
	for(auto x : tmp){
		now += x;
		vec.push_back(now);
	}
}
inline void solve(){
	const ll a = io.readll(),b = io.readll();
	if(a == b){
		io.write(a,'\n');
		return;
	}
	resolve(veca,a);
	resolve(vecb,b);
	// for(auto x : veca)io.write(x,' ');
	// io.lf();
	// for(auto x : vecb)io.write(x,' ');
	// io.lf();
	for(unsigned int i = 0;;i++){
		if(veca[i + 1] != vecb[i + 1] || (i + 1 == veca.size()) || (i + 1 == vecb.size())){
			io.write(veca[i],'\n');
			return;
		}
	}
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("fafa.in","r",stdin);
#endif
	const int m = io.readi();
	for(int i = 1;i <= m;i++)solve();
	return 0;
}

标签：int,题解,void,斐波,io,inline,那契,now,obuf
来源： https://www.cnblogs.com/colazcy/p/14031975.html