本题和 [96.不同的二叉搜索树](https://www.programmercarl.com/0096.%E4%B8%8D%E5%90%8C%E7%9A%84%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html) 比较像

* 取模这里很容易出错 
* 过程中所用到的数值都有可能超过int，所以要改用longlong 

```CPP 
#include <iostream>
#include <vector>
using namespace std;

long long mod = 1e9 + 7;
long long dp(int t, vector<long long>& memory) {
    if (t % 2 == 0) return 0;
    if (t == 1) return 1;
    if (memory[t] != -1) return memory[t];

    long long result = 0;
    // 枚举左右子树节点的数量
    for (int i = 1; i < t; i += 2) {
        long long leftNum = dp(i, memory); // 左子树节点数量为i
        long long rightNum = dp(t - i - 1, memory); // 右子树节点数量为t - i - 1
        result += (leftNum * rightNum) % mod; // 注意这里是乘的关系
        result %= mod;
    }
    memory[t] = result;
    return result;
}
int main() {
    int n;
    cin >> n;
    vector<long long> memory(n + 1, -1);
    cout << dp(n,  memory) << endl;
}
```


```CPP
#include <iostream>
#include <vector>
#include <cstring>

using namespace std;

const int MOD = 1000000007;

int main() {
    int num;
    cin >> num;

    if (num % 2 == 0) {
        cout << 0 << endl;
        return 0;
    }

    vector<long long> dp(num + 1, 0);
    dp[1] = 1;

    for (int i = 3; i <= num; i += 2) {
        for (int j = 1; j <= i - 2; j += 2) {
            dp[i] = (dp[i] + dp[j] * dp[i - 1 - j]) % MOD;
        }
    }

    cout << dp[num] << endl;
    return 0;
}

```


第二题的代码 

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

long fastexp(long base,long n,long mod){
    long answer = 1;
    while(n > 0){
        if(n % 2 == 1){
            answer = (answer * base) % mod;
        }
        base = (base * base) % mod;
        n /= 2;
    }
    return answer;
}
int kawaiiStrings(int n) {
    // write code here
    std::vector<long> f(n + 1), g(n + 1), h(n + 1);
    long mod = 1000000007;
    for (long i = 2; i <= n; i++) g[i] = (g[i - 1] * 2 + (i - 1) * fastexp(2,i-2,mod)) % mod;
    for (long i = 3; i <= n; i++) f[i] = ((f[i - 1] * 3) % mod + g[i - 1]) % mod;
    for (long i = 3; i <= n; i++) h[i] = (fastexp(3, i - 3, mod) + h[i - 1] * 3 - h[i - 3]) % mod;
    return (f[n]-h[n]+mod)%mod;

}

int main(){
    int n;
    cin >> n;
    cout << kawaiiStrings(n) << endl;
    return 0;
}