Submission #891943

#	Time	Username	Problem	Language	Result	Execution time	Memory
891943		josanneo22	Kangaroo (CEOI16_kangaroo)	C++17	0 / 100	1 ms	348 KiB

kangaroo

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;

const int mod = 1E9 + 7;
const int _N = 2500;
i64 dp[_N][_N];
/*
    dp[i][j] = 已经排了i个数,连通块=j
    放入某个块: dp[i][j]=dp[i-1][j]*j
    新加一个块: dp[i][j]=dp[i-1][j-1]*j
    合并两个块: dp[i][j]=dp[i-1][j+1]*j
    若i==A || i==B, 我们不可以用来合并而且也不能随便加入块:
        dp[i][j] = dp[i-1][j]
        dp[i][j] = dp[i-1][j-1]
*/
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int N, A, B; cin >> N >> A >> B;
    dp[1][1] = 1;
    for (int i = 2; i <= N; i++) {
        for (int j = 1; j <= N; j++) {
            dp[i][j] = 0;
            if (i == A || i == B) {
                dp[i][j] += dp[i - 1][j];
                dp[i][j] %= mod;
                dp[i][j] += dp[i - 1][j - 1];
                dp[i][j] %= mod;
                continue;
            }
            int cnt = 2;
            dp[i][j] += dp[i - 1][j - 1] * (j - cnt);//不能放头，不能放尾
            dp[i][j] %= mod;
            dp[i][j] += dp[i - 1][j + 1] * j;
            dp[i][j] %= mod;
        }
    }
    cout << dp[N][1] << '\n';
}

• Subtask 1: 0 / 6
• Subtask 2: 0 / 30
• Subtask 3: 0 / 15
• Subtask 4: 0 / 49