三种基因，我们分别列出其生成函数：
$$\begin{gathered} F(x)=\sum _{n\ge 0}\frac{x^n}{n!}=e^x \\ G(x)=\sum _{n\ge 0}\frac{x^{2n+1}}{(2n+1)!}=\frac{1}{2}\left(\sum _{n\ge 0}\frac{x^n}{n!}-\sum _{n\ge 0}(-1{)}^n\frac{x^n}{n!}\right)=\frac{1}{2}\left(e^x-e^{-x}\right) \\ H(x)=\sum _{n\ge 0}\frac{x^{2n}}{(2n)!}=\frac{1}{2}\left(\sum _{n\ge 0}\frac{x^n}{n!}+\sum _{n\ge 0}(-1{)}^n\frac{x^n}{n!}\right)=\frac{1}{2}\left(e^x+e^{-x}\right) \end{gathered}$$

$$\begin{gathered} A(x)={\left(F(x)G(x)H(x)\right)}^4=\frac{1}{256}\left(e^{12x}-4e^{8x}+6^{4x}+e^{-4x}-4\right) \\ 对上面进行泰勒展开得到第n项的系数再乘上n!就是答案了 \\ ans[n]=\frac{1}{256}\left(12^n-4\times 8^n+6\times 4^n+(-4{)}^n\right) \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int mod = 1e9, phi = 1e8;int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);long long n;while (scanf("%lld", &n) && n) {if (n < 4) {puts("0");}else if (n < phi) {printf("%lld\n", (1ll * 81 * quick_pow(12, n - 4) % mod - quick_pow(8, n - 2) + 1ll *  6 * quick_pow(4, n - 4) + quick_pow(-4, n - 4) + mod) % mod);}else {int ans = 1ll * 81 * quick_pow(12, (n - 4) % phi + phi) % mod;ans = (ans - quick_pow(8, (n - 2) % phi + phi) + mod) % mod;ans = (ans + 1ll * 6 * quick_pow(4, (n - 4) % phi + phi) % mod) % mod;int res = quick_pow(4,(n - 4) % phi + phi);if((n - 4) & 1){ans = (ans - res + mod) % mod;}else {ans = (ans + res) % mod;}printf("%d\n", ans);}}return 0;
}