제출 #1293854

#제출 시각아이디문제언어결과실행 시간메모리
1293854darkdevilvaqif캥거루 (CEOI16_kangaroo)C++20
0 / 100
1 ms332 KiB
#pragma GCC optimize("O3")
#pragma GCC optimize ("unroll-loops")
// #pragma GCC target("avx2")
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define ld long double
#define all(v, l) v.begin() + l, v.end()
#define rall(v, l) v.rbegin(), v.rend() - l
#define pb push_back
#define pf push_front
#define rsz resize
#define fi first
#define se second
#define LMAX LLONG_MAX
#define LMIN LLONG_MIN
#define IMAX INT_MAX
#define IMIN INT_MIN
#define endl "\n"
#define newline cout << endl;
using namespace std;
const ll MOD = 1000000007;
 
// structs

// globals

// variables
int n, s, e;

// iterators
int i, k;

// notes
/*
 -stuff you should look for-
* int overflow, array bounds
* special cases (n=1?)
* do something instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH

continue - skip the rest in the loop
dp[a][b][c] represents the number of ways to come to path a in b turns, jumping (c == 0 ? down : up)
k - turn
*/

// functions
ll GCD(ll numeroune, ll numerodeux);
ll LCM(ll numeroune, ll numerodeux);
ll power(ll numeroune, ll numerodeux);
 
void solve()
{
    cin >> n >> s >> e;
    
    vector <vector <vector <ll> > > dp(n + 1, vector <vector <ll> > (n + 1, vector <ll> (2, 0LL)));
    vector <ll> pf0(n + 1), pf1(n + 1);
    dp[s][1][1] = dp[s][1][0] = 1LL;
    for (k = 2; k <= n; k++)
    {
        for (i = 1; i <= n; i++)
        {
            pf0[i] = (pf0[i - 1] + dp[i][k - 1][0]) % MOD;
            pf1[i] = (pf1[i - 1] + dp[i][k - 1][1]) % MOD;
        }
        
        for (i = 1; i <= n; i++)
        {
            dp[i][k][0] = pf1[i - 1];
            dp[i][k][1] = (pf0[n] - pf0[i]) % MOD;
        }
    }
    
    cout << (dp[e][n][0] + dp[e][n][1]) % MOD;
}
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
        newline
    }
}
 
ll GCD(ll numeroune, ll numerodeux)
{
    if (!numeroune)
    {
        return numerodeux;
    }
    return GCD(numerodeux % numeroune, numeroune);
}

ll LCM(ll numeroune, ll numerodeux)
{
    return numeroune * numerodeux / GCD(numeroune, numerodeux);
}

ll power(ll numeroune, ll numerodeux)
{
    ll res = 1;
    while (numerodeux) 
    {
        if (numerodeux & 1)
        { 
            res *= numeroune;
        }
        numeroune *= numeroune;
        numerodeux >>= 1;    
    }
    return res;
}

/*
$$$$$$$$\ $$$$$$$$\ 
$$  _____|\____$$  |
$$ |          $$  / 
$$$$$\       $$  /  
$$  __|     $$  /   
$$ |       $$  /    
$$$$$$$$\ $$$$$$$$\ 
\________|\________|
*/
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