Submission #869438

#	Time	Username	Problem	Language	Result	Execution time	Memory
869438		sleepntsheep	Kangaroo (CEOI16_kangaroo)	C++17	0 / 100	0 ms	348 KiB

kangaroo

#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <deque>
#include <set>
#include <utility>
#include <array>

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx2,tune=native")


using namespace std;
#define ALL(x) x.begin(), x.end()
#define ShinLena cin.tie(nullptr)->sync_with_stdio(false);
using ll = long long;
#define N 2005
const ll M = 1000000007;

int n, cs, cf, dp[2][N][2][2];

int main()
{
    ShinLena;
    cin >> n >> cs >> cf;
    dp[0][0][0][0] = 1;
    for (int I = 1, i = 1; i <= n; ++i, I ^= 1)
    {
        memset(dp[I], 0, sizeof *dp);
        if (i == cs)
        {
            for (int j = 1; j <= i; ++j)
            {
                dp[I][j][0][0] = 0;
                dp[I][j][1][0] = dp[!I][j][0][0] % M;
                dp[I][j][1][1] = (dp[!I][j-1][0][1] + dp[!I][j][0][1]) % M;
            }
        }
        if (i == cf)
        {
            for (int j = 1; j <= i; ++j)
            {
                dp[I][j][0][1] = (dp[!I][j-1][0][0] + dp[!I][j][0][0]) % M;
                dp[I][j][1][1] = (dp[!I][j-1][1][0] + dp[!I][j][1][0]) % M;
            }
        }
        else
        {
            for (int j = 1; j <= i; ++j)
            {
                dp[I][j][0][0] = 
                    ((1ll * dp[!I][j+1][0][0] * j) % M
                    + (1ll * dp[!I][j-1][0][0] * (j+1) % M)
                    ) % M;
                
                dp[I][j][1][0] =
                    (
                     (1ll * dp[!I][j][0][0]) + 
                     (1ll * dp[!I][j+1][1][0] * j) % M
                    ) % M;

                dp[I][j][0][1] =
                    (
                     (1ll * dp[!I][j][0][0]) + 
                     (1ll * dp[!I][j+1][0][1] * j) % M
                    ) % M;

                dp[I][j][1][1] =
                    (
                     (1ll * dp[!I][j][0][1] + dp[!I][j][1][0]) % M + 
                     (1ll * dp[!I][j+1][1][1] * j) % M
                    ) % M;

            }
        }
    }
    cout << dp[n&1][1][1][1];
    return 0;
}

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