# 2749 피보나치 수 3

```cpp
#include"2749.h"
using namespace std;
const int MOD = 1000000;
typedef long long ll;

map<ll, int> dict;

int Fibo_L(int n)
{
	if (n == 0) return 0;
	if (n == 1) return 1;
	else return Fibo_L(n - 1) + Fibo_L(n - 2);
}

ll Fibo(ll n)
{
	if (n <= 3) return Fibo_L(n);
	if (dict.count(n)) return dict[n];
	ll res;
	if (n % 2 == 0) // 짝수일 경우
	{
		ll a = Fibo(n / 2) % MOD;
		ll b = Fibo(n / 2 + 1) % MOD;
		ll c = Fibo(n / 2 - 1) % MOD;
		res = a * (b + c) % MOD;
	}
	else // 홀수일 경우
	{
		ll a = Fibo((n + 3) / 2) % MOD;
		ll b = Fibo((n - 1) / 2) % MOD;
		ll c = Fibo((n + 1) / 2) % MOD;
		ll d = Fibo((n - 3) / 2) % MOD;
		res = (a*b + c*d) % MOD;
	}
	dict[n] = res;
	return res;
}


void B2749::Solution()
{
	long long n;
	cin >> n;
	cout << Fibo(n);
}
```

실수

* `n<=3` 조건을 `<3`으로 걸어서 3이면 무한히 안 끝났었음
* 점화식 전개할 때 각각 F\_2, F\_1으로 놔야 하는데 둘 다 F\_1으로 놔버려서 식 잘못 정리됐었음
* 캐싱 도입 안 했더니 계산량이 지수적으로 증가해서 시간 초과.

```cpp
#include <iostream>
using namespace std;
long long arr[2000001];
int main() {
    arr[0] = 0;
    arr[1] = 1;
    long long N;
    cin>>N;
    N%=1500000;
    for(int i=2;i<=N;i++)
        arr[i]=(arr[i-1]+arr[i-2])%1000000;
    cout<<arr[N];
}
```

피사노 주기를 이용하면 어처구니 없게 짧게 풀 수 있다. 속도는 약간 더 느리긴 함.


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