제출 #1294240

#제출 시각아이디문제언어결과실행 시간메모리
1294240xosqedemrufo캥거루 (CEOI16_kangaroo)C++20
100 / 100
61 ms55168 KiB
//Author RufatM
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<string> vs;
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl '\n'
#define pb push_back
#define pf push_front
#define eb emplace_back
#define ff first
#define ss second
#define all(x) begin(x),end(x)
#define rall(x) rbegin(x),rend(x)
#define mt19937_64 mt_rand(chrono::steady_clock::now().time_since_epoch().count())
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
const int MOD=1000000007;
const int INF=1000000007;
const ll LINF=4e18;
const int MAXN=2005;
int n,s,l;
ll dp[MAXN][MAXN][3];
void modpow(ll &a,ll b){
    a+=b;
    if(a>=MOD){
        a-=MOD;
    }
    if(a<0){
        a+=MOD;
    }
}
signed main(){
    fastio;
    #ifndef ONLINE_JUDGE
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    #endif
    int t=1;
    //cin >> t;
    while(t--){
        cin >> n >> s >> l;
        if(s==1 or l==1){
            dp[1][1][1] = 1;
        }
        else{
            dp[1][1][0] = 1;
        }
        for(int i=2;i<=n;i++){
            for(int j=1;j<i;j++){
                if(i==s or i==l){
                    for(int k=0;k<2;k++){
                        modpow(dp[i][j+1][k+1],dp[i-1][j][k]);
                        modpow(dp[i][j][k+1],dp[i-1][j][k]);
                    }
                }
                else{
                    for(int k=0;k<3;k++){
                        modpow(dp[i][j+1][k],(dp[i-1][j][k]*(j+1-k))%MOD);
                        modpow(dp[i][j-1][k],(dp[i-1][j][k]*(j-1))%MOD);
                    }
                }
            }
        }
        cout << dp[n][1][2] % MOD << endl;
    }
}
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