Reading comprehension

题意:Fi满足:if (i & 1)ans = (ans * 2 + 1) % m; else ans = ans * 2 % m;,给出n,m，问Fn为多少。

当i为奇数时,Fi=4F(i-1)+1,当j为偶数时，Fj=4F(j-1)+2.我们可以得到这样一个关系：

#include
#include
#include
#include

using namespace std;

typedef long long ll;

const int N = 2e5 + 10;

int arr[N];

ll mod;

struct matrix {
	ll rix[5][5];
}ans,test;

ll quick_pow(ll a,ll b) {
	ll ans = 1;
	a %= mod;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}

matrix mul(int n,matrix rix1,matrix rix2) {
	matrix temp;
	memset(temp.rix, 0, sizeof(temp.rix));
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				temp.rix[i][j] = (temp.rix[i][j] + rix1.rix[i][k] * rix2.rix[k][j])%mod;
			}
		}
	}
	return temp;
}

void quick_rix(int n,int k) {
	memset(ans.rix, 0, sizeof(ans.rix));
	for (int i = 1; i <= n; i++) ans.rix[i][i] = 1;
	while (k) {
		if (k & 1) ans = mul(n,ans,test);
		test = mul(n, test, test);
		k >>= 1;
	}
	//return ans;
}

int main() {
	ios::sync_with_stdio(false);
	matrix rax;
	int n;
	while (cin >> n >> mod) {
		memset(rax.rix, 0, sizeof(rax.rix));
		memset(test.rix, 0, sizeof(test.rix));
		rax.rix[1][1] = 1;
		rax.rix[2][1] = 2;
		rax.rix[1][2] = 1;
		rax.rix[2][2] = 2;
		test.rix[1][1] = 4;
		test.rix[2][1] = 1;
		test.rix[1][2] = 0;
		test.rix[2][2] = 1;
		if (n == 1) {
			cout << (rax.rix[1][1] + mod) % mod << endl;
		}
		else if (n == 2) {
			cout << (rax.rix[2][1] + mod) % mod << endl;
		}
		else {
			quick_rix(2, (n - 1) / 2);
			ans = mul(2, rax, ans);
			if (n & 1) {
				cout << (ans.rix[1][1] + mod) % mod << endl;
			}
			else {
				cout << (ans.rix[2][1] + mod) % mod << endl;
			}
		}
	}
}