題目大意：問長度為$n$的$Sam$數有幾個，$Sam$數的定義為沒有前導零，相鄰兩個數字之差絕對值小於等於$2$的數

題解：發現轉移方程一定，可以矩陣快速冪。

卡點：沒有特判$n=1$的情況

C++ Code：

#include <cstdio>
const int mod = 1e9 + 7;
inline int abs(int a) {return a < 0 ? -a : a;}
inline void up(int &a, int b) {a += b - mod; a += a >> 31 & mod;}
struct matrix {
	#define M 10
	int s[M][M];
	inline void E() {
		for (int i = 0; i < M; i++) s[i][i] = 1;
	}
	inline void init() {
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) s[i][j] = abs(i - j) <= 2;
		}
	}
	inline friend matrix operator * (const matrix &lhs, const matrix &rhs) {
		matrix res;
		long long ans;
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) {
				ans = 0;
				for (int k = 0; k < M; k++) ans += static_cast<long long> (lhs.s[i][k]) * rhs.s[k][j];
				res.s[i][j] = ans % mod;
			}
		}
		return res;
	}
	#undef M
} ans, base;

long long n;
int main() {
	scanf("%lld", &n);
	if (n == 1) {
		puts("10");
		return 0;
	}
	base.init();
	for (int i = 1; i < 10; i++) ans.s[0][i] = 1;
	for (n--; n; n >>= 1, base = base * base) if (n & 1) ans = ans * base;
	int res = 0;
	for (int i = 0; i < 10; i++) up(res, ans.s[0][i]);
	printf("%d\n", res);
	return 0;
}