ural1017 Staircases (动态规划)

设dp[i][j]表示使用i个砖块，最后一个台阶的方块数量为j的不同方案数（先不考虑条件Every staircase consists of at least two steps）

那么状态转移方程即为dp[i][j] = ∑dp[i-j][k]  (0 <= k <= min{i - j, j - 1})

最后考虑每个楼梯都必须包含至少两层，这时只需要最终答案减1便可

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 505;LL dp[N][N];int main() {    int n; scanf("%d", &n);    memset(dp, 0, sizeof(dp));    dp[0][0] = 1;    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= i; j++) {            for (int k = 0; k <= min(i - j, j - 1); k++) {                dp[i][j] += dp[i-j][k];            }        }    }    LL ans = 0;    for (int i = 0; i <= n; i++)        ans += dp[n][i];    printf("%lld\n", ans - 1);    return 0;}