学步园

大意不再赘述。

思路：由于我怕精度丢失，所以在1~30内的Fibonacci，我都直接算出来，大于30的用通项公式算。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;

typedef unsigned long long ULL;

ULL f[31];

int n;

void init()
{
	f[1] = 1, f[2] = 1;
	for(int i = 3; i <= 31; i++) f[i] = f[i-1]+f[i-2];
}

ULL cal(int n)
{
	double f = (1.0+sqrt(5.0))/2.0;
	double a = -0.5*log10(5.0)+((double)n)*log10(f);
	ULL ans = (ULL)a + 1;
	return ans;
}

void solve()
{
	if(n <= 30)
	{
		double ans = log10(f[n])+1.0;
		printf("%llu %llu\n", (ULL)ans, f[n]);
	}
	else printf("%llu\n", cal(n));
}

int main()
{
	init();
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d", &n);
		solve();
	}
	return 0;
}