제출 #172209

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
172209		vex	캥거루 (CEOI16_kangaroo)	C++14	51 / 100	168 ms	136340 KiB

kangaroo

#include<bits/stdc++.h>
#define pii pair<int,int>
#define ll long long
#define pll pair<ll,ll>
#define pb push_back
#define maxn 1000005
#define mod 1000000007
#define val first
#define poj second
using namespace std;

int n;
int poc,kraj;
ll dp[205][205][205][2];

void calc()
{
    for(int duz=2;duz<=n;duz++)
    {
        //cout<<endl<<endl;
        //cout<<duz<<endl;
        for(int i=1;i<=duz;i++)
        {
            for(int j=i+1;j<=duz;j++)
            {
                if(i==1)
                {
                    dp[duz][i][j][0] = 0;
                    if(duz==2)
                    {
                        dp[duz][i][j][1] = 1;
                    }
                }
                else
                {
                    dp[duz][i][j][0] = dp[duz][i-1][j][0] + dp[duz - 1][i-1][j-1][1];
                    dp[duz][i][j][0] %= mod;
                }

            }
        }

        for(int i=duz;i>0;i--)
        {
            for(int j=i+1;j<=duz;j++)
            {
                if(i + 1 == j)
                {
                    for(int k=j;k<=duz-1;k++)
                    {
                        dp[duz][i][j][1] += dp[duz-1][j-1][k][duz%2];
                        dp[duz][i][j][1] %= mod;
                    }
                }
                else
                {
                    dp[duz][i][j][1] = dp[duz][i+1][j][1] + dp[duz - 1][i][j-1][0];
                    dp[duz][i][j][1] %= mod;
                }
            }
        }

        /*for(int i=1;i<=duz;i++)
        {
            cout<<i<<": ";
            for(int j=i+1;j<=duz;j++)
            {
                cout<<j<<","<<dp[duz][i][j][0]<<","<<dp[duz][i][j][1]<<" ";
            }
            cout<<endl;
        }*/
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    cin>>n>>poc>>kraj;
    if(poc > kraj)swap(poc,kraj);
    calc();
    cout<<(dp[n][poc][kraj][0] + dp[n][poc][kraj][1])%mod<<endl;

    return 0;
}

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