P3986 斐波那契数列

傻逼题。

首先，\(Fib_{47}=2.971215073\times10^9\)。

所以说我们把这个系数整出来，然后一个一个验就可以了。

这里用 Ex-gcd 就可以了，找到通解，然后算出正整数解的个数就可以了。

//#pragma GCC optimize(2)
#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <stack>
#include <tuple>
#define ll long long
#define fp(a,b,c) for(ll a=b;a<=c;a++)
#define fd(a,b,c) for(ll a=b;a>=c;a--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define inf 0x3f3f3f3f
#define int ll
#define base 127
#define mod 1000000007
#define cel(a,b) (a%b)?(a/b+1):a/b 

using namespace std;

inline int rd(){
	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;}
inline ll lrd(){
	ll 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;}

int exgcd(int &x,int &y,int a,int b){
	if(!b){
		x=1,y=0;return a;
	}
	int sum=exgcd(x,y,b,a%b);
	int t=x;
	x=y;
	y=t-a/b*y;
	return sum;
}

int k;
pii xs[100];

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0); 
	k=rd();
	xs[1]={1,0};
	xs[2]={0,1};
	fp(i,3,47){
		xs[i].first=xs[i-1].first+xs[i-2].first;
		xs[i].second=xs[i-1].second+xs[i-2].second;
	}
	ll ans=0;
	fp(i,3,47){
		int x,y;
		int gd=exgcd(x,y,xs[i].first,xs[i].second);
		int a=xs[i].first,b=xs[i].second;
		if(k%gd!=0) continue ;
		x*=(k/gd),y*=(k/gd);
		x=(x%b+b)%b;
		if(x==0)
			x=b;
		y=(k-(a*x))/b;
		if(y<0) continue ;
		(ans+=cel(y,a))%=mod;
	}
	cout << ans << endl;
	return 0;
}