參考資訊:
https://www.cnblogs.com/grandyang/p/9179556.html
題目:
解答:
int numTilings(int n) { int cc = 0; long dp[1000] = { 1, 1, 2 }; long mod = 1e9 + 7; for (cc = 3; cc <= n; cc++) { dp[cc] = ((dp[cc - 1] * 2) + dp[cc - 3]) % mod; } return dp[n]; }