思路：dp(i)表示第i天的猪的数量，g(i)表示第i天新出生的猪的数量，d(i) = d(i-1) * 2 - g(i-2)， g(i) = d(i-1)

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 20 + 5;
int ans[maxn], g[maxn];
void init() {
	ans[0] = g[0] = 0;
	ans[1] = g[1] = 1;
	for(int i = 2; i <= 20; ++i) {
		g[i] = ans[i-1];
		ans[i] = g[i] * 2;
		if(i-2 >= 0) ans[i] -= g[i-2];
	}
}
int main() {
	init();
	int T, n;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		printf("%d\n", ans[n]);
	}
	return 0;
}

如有不当之处欢迎指出！