Submission #756068

#	Time	Username	Problem	Language	Result	Execution time	Memory
756068		browntoad	Kangaroo (CEOI16_kangaroo)	C++14	100 / 100	21 ms	22996 KiB

kangaroo

#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast", "unroll-loops")
using namespace std;
#define ll long long
#define int ll
#define FOR(i,a,b) for (int i = (a); i<(b); i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1; i>=0; i--)
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define SQ(x) (x)*(x)
#define pii pair<int, int>
#define pip pair<int, pii>
#define pdd pair<double ,double>
#define pcc pair<char, char>
#define endl '\n'
//#define TOAD
#ifdef TOAD
#define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl
#define IOS()
#else
#define bug(...)
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#endif

const ll inf = 1ll<<60;
const ll iinf = 2147483647;
const ll mod = 1e9+7;
const ll maxn = 2005;
const ll maxm = 2005;
const double PI=acos(-1);

ll pw(ll x, ll p, ll m=mod){
    ll ret=1;
    while (p>0){
        if (p&1){
            ret*=x;
            ret%=m;
        }
        x*=x;
        x%=m;
        p>>=1;
    }
    return ret;
}

ll inv(ll a, ll m=mod){
    return pw(a,m-2,m);
}
int dp[maxn][maxn];
signed main(){
    IOS();
    int n, cs, cf; cin>>n>>cs>>cf;
    if (cs > cf) swap(cs, cf);
    dp[0][0] = 1;
    REP1(i, n){
        REP1(j, i){
            if (i < cs){
                dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1]*(j+1)%mod*j%mod;
            }
            if (i == cs){
                dp[i][j] = dp[i-1][j-1] + dp[i-1][j]*j%mod;
            }
            if (i > cs && i < cf){
                dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1]*j%mod*j%mod; // j * (j-1) + 1 * j
            }
            if (i == cf){
                if (i == n && j == 1) dp[i][j] = dp[i-1][j];
                else dp[i][j] = dp[i-1][j-1] + dp[i-1][j]*(j-1)%mod;
            }
            if (i > cf){
                if (i == n && j == 1) dp[i][j] = dp[i-1][j+1];
                else dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1]*(j-1)%mod*j%mod;
            }
            dp[i][j] %= mod;
        }
    }
    cout<<dp[n][1]<<endl;
}

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