CF1105C Ayoub and Lost Array

\(\large{题目链接}\)
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\(\Large\textbf{Solution: } \large{这道题我一开始以为是推式子直接算，然后想了好久都没想出来。查看标签，发现是dp。\\首先预处理出区间中膜3余1、2、0的数有多少个，因为我们只关心余数。设f_{i,j}表示到第i个数字余数为j的方案个数，那么直接进行转移即可。时间复杂度\text{O(n)}。\\查看题解，发现可以矩阵加速递推，我还是太菜了，没有想到。\\设a是膜3余数为0的个数，b是余1，c是余2，那么可以写成这么一个矩乘，直接快速幂即可。时间复杂度\text{O(logn)}。\\\begin{bmatrix} f_{0} & f_{1} & f_{2} \end{bmatrix} \times \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix}}\)
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\(\Large\textbf{Summary: } \large{1.并不是所有数学题都能推式子，有些也与dp相关。\\2.对于递推，如果能找到一个固定关系，可以考虑矩阵加速。}\)
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\(\Large\textbf{Code: }\)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int p = 1e9 + 7;
const int N = 2e5 + 5;

ll n, f[N][3];

int main() {
	ios::sync_with_stdio(false);
	ll l, r;
	cin >> n >> l >> r;
	ll a, b, c;
	a = r / 3 - (l - 1) / 3;
	b = (r + 2) / 3 - (l + 1) / 3;
	c = (r + 1) / 3 - l / 3;
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		f[i][0] = ((f[i - 1][0] * a) % p + (f[i - 1][1] * c) % p + (f[i - 1][2] * b) % p) % p;
		f[i][1] = ((f[i - 1][0] * b) % p + (f[i - 1][1] * a) % p + (f[i - 1][2] * c) % p) % p;
		f[i][2] = ((f[i - 1][0] * c) % p + (f[i - 1][1] * b) % p + (f[i - 1][2] * a) % p) % p;
	}
	cout << f[n][0] << endl;
	return 0;
}

\(\Large\textbf{Code: }\)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 2e5 + 5;
const int p = 1e9 + 7;

int n;

struct juz {
	int a[5][5];
	juz () {
		memset(a, 0, sizeof (a));
	}
};

juz operator * (const juz x, const juz y) {
	juz z;
	for (int k = 1; k <= 3; ++k)
		for (int i = 1; i <= 3; ++i)
			for (int j = 1; j <= 3; ++j)
				z.a[i][j] = (1ll * z.a[i][j] + 1ll * x.a[i][k] * y.a[k][j] % p) % p;
	return z;
}

int main() {
	ios::sync_with_stdio(false);
	int l, r;
	cin >> n >> l >> r;
	int a, b, c;
	a = r / 3 - (l - 1) / 3;
	b = (r + 2) / 3 - (l + 1) / 3;
	c = (r + 1) / 3 - l / 3;
	juz m, ans;
	ans.a[1][1] = 1;
	m.a[1][1] = m.a[2][2] = m.a[3][3] = a;
	m.a[1][2] = m.a[2][3] = m.a[3][1] = b;
	m.a[1][3] = m.a[2][1] = m.a[3][2] = c;
	for (; n ; n >>= 1, m = m * m) if (n & 1) ans = ans * m;
	cout << ans.a[1][1];
	return 0;
}