用斐波那契分解正整数

分解正整数"/>

用斐波那契分解正整数

观察这个形式，如果交替做，就是个斐波那契数列

打表可得，任何正整数都可以大约由 log ⁡ \log log 个斐波那契数加起来

然后直接拼斐波那契数即可

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else#define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
//#define N
//#define M
//#define mo
int n, m, i, j, k, T;
int f[100], mx, x, y; 
vector<pair<int, int> >v; 
vector<int>G[100]; void suan(int op) {if(op==1) ++x; if(op==2) ++y; if(op==3) x+=y; if(op==4) y+=x; debug("[%lld %lld]\n", x, y); 
}void dfs(int x) {if(!x) {for(auto t : v) {debug("%lld %lld | %lld\n", mx, t.se, t.fi); if(t.se%2) G[mx-t.se].pb(2); else G[mx-t.se].pb(1); }for(i=k=0; i<=mx; ++i) k+=G[i].size()+1; printf("%lld\n", k); for(i=0; i<=mx; ++i) {for(auto j : G[i]) {printf("%lld\n", j); suan(j); }printf("%lld\n", (i%2 ? 3 : 4)); suan((i%2 ? 3 : 4)); }return ; }for(int i=90; i>=1; --i)if(x>=f[i]) {mx=max(mx, i); v.pb({f[i], i}); dfs(x-f[i]); return ; }
}signed main()
{#ifdef LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif
//	T=read();
//	while(T--) {
//
//	}n=read(); f[0]=f[1]=1; for(i=2; i<=90; ++i) f[i]=f[i-1]+f[i-2]; dfs(n); return 0;
}