Sicily 1121. Tri Tiling

#include <iostream>

using namespace std;

long long d[31];

void dp() {
for (int i = 4; i < 31; i += 2) {
d[i] = 3 * d[i - 2];	// i)3个块组成独立矩形
// ii)不可垂直分割，作为整体，方案数累加
for (int j = 0; j <= i - 4; j += 2)
d[i] += 2 * d[j];
}
}

int main(int argc, char **argv) {
int n;
// 递推起点
d[0] = 1;
d[2] = 3;

while (cin >> n && n != -1) {
if (n % 2)
cout << 0 << endl;
else {
dp();
cout << d
<< endl;
}
}
return 0;
}

#include <iostream>

using namespace std;

long long d[31];

int main(int argc, char **argv) {
int n;
// 解递推式f(n) = 3f(n-2) + 2(f(n-4) + f(n-6) + ... + f(0))得f(n) = 4f(n-2) - f(n-4)， 思路更清晰
d[0] = 1;
d[2] = 3;
for (int i = 4; i < 31; ++i)
d[i] = 4 * d[i - 2] - d[i - 4];

while (cin >> n && n != -1) {
if (n % 2)
cout << 0 << endl;
else
cout << d
<< endl;

}
return 0;
}