Sicily 1121. Tri Tiling

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#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[n] << 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;elsecout << d[n] << endl;}return 0;}

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