HackerRank Unique Divide And Conquer problem solution

In this HackerRank Unique Divide And Conquer problem solution we have Given the number of vertices N, count the number of tree T’s such that the Divide-and-Conquer approach works determinately on them. As this number can be quite large, your answer must be modulo M.

HackerRank Unique Divide And Conquer problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {
    FastScanner in;
    PrintWriter out;

    void solve() {
        int n = in.nextInt();
        int p = in.nextInt();
        int[][] c = new int[n + 1][n + 1];
        c[0][0] = 1;
        for (int i = 1; i <= n; i++) {
            c[i][0] = 1;
            for (int j = 1; j <= n; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
                if (c[i][j] >= p) {
                    c[i][j] -= p;
                }
            }
        }
        long[] dpSum = new long[n + 1];
        long[] ans = new long[n + 1];
        long[] dpBad = new long[n + 1];
        dpSum[0] = 1;
        for (int sz = 1; sz <= n; sz++) {
            for (int big = (1 + sz) / 2; big < sz; big++) {
                dpBad[sz] += dpSum[sz - 1 - big] * ans[big] % p * big % p
                        * c[sz - 1][big] % p;
            }
            dpBad[sz] %= p;
            ans[sz] = (dpSum[sz - 1] - dpBad[sz] + p) * sz % p;
            for (int size1 = 1; size1 <= sz; size1++) {
                dpSum[sz] += ans[size1] * size1 % p * c[sz - 1][size1 - 1] % p
                        * dpSum[sz - size1] % p;
            }
            dpSum[sz] %= p;
        }
        out.println(ans[n]);
    }

    void run() {
        try {
            in = new FastScanner(new File("object.in"));
            out = new PrintWriter(new File("object.out"));

            solve();

            out.close();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

    void runIO() {

        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        public FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return null;
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    public static void main(String[] args) {
        new Solution().runIO();
    }
}

Unique Divide And Conquer problem solution in C++.

#include <iostream>
#include <fstream>
#include <sstream>

#include <vector>
#include <set>
#include <bitset>
#include <map>
#include <deque>
#include <string>

#include <algorithm>
#include <numeric>

#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>

#define pb push_back
#define pbk pop_back
#define mp make_pair
#define fs first
#define sc second
#define all(x) (x).begin(), (x).end()
#define foreach(i, a) for (__typeof((a).begin()) i = (a).begin(); i != (a).end(); ++i)
#define len(a) ((int) (a).size())

#ifdef CUTEBMAING
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define eprintf(...) 42
#endif

using namespace std;

typedef long long int64;
typedef long double ld;
typedef unsigned long long lint;

const int inf = (1 << 30) - 1;
const int64 linf = (1ll << 62) - 1;
const int N = 1e4 + 100;
int M;

inline int power(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) {
            res = (1ll * res * a) % M;
        }
        a = (1ll * a * a) % M;
        b >>= 1;
    }
    return res;
}

inline int inv(int x) {
    return power(x, M - 2);
}

int n;
int fact[N], ifact[N];

void precalc() {
    fact[0] = ifact[0] = 1;
    for (int i = 1; i <= n; i++) {
        fact[i] = (1ll * fact[i - 1] * i) % M;
        ifact[i] = (1ll * ifact[i - 1] * inv(i)) % M;
    }
}

int f[N], dp[N];

int main() {
    cin >> n >> M;
    precalc();
    f[1] = 1;
    dp[0] = dp[1] = 1;

    for (int i = 2; i <= n; i++) {
        f[i] = dp[i - 1];
        for (int j = (i + 1) / 2; j <= i - 1; j++) {
            int cur1 = (1ll * f[j] * fact[i - 1]) % M;
           
            int cur2 = (1ll * cur1 * ifact[j]) % M;
            int cur3 = (1ll * cur2 * ifact[i - 1 - j]) % M;
            int cur4 = (1ll * cur3 * j) % M;
            f[i] -= (1ll * cur4 * dp[i - j - 1]) % M;
            if (f[i] < 0) {
                f[i] += M;
            }
        }
        f[i] = (1ll * f[i] * i) % M;
        for (int j = 1; j <= i; j++) {
            int cur1 = (1ll * f[j] * fact[i - 1]) % M;
            int cur2 = (1ll * cur1 * ifact[j - 1]) % M;
            int cur3 = (1ll * cur2 * ifact[i - j]) % M;
            int cur4 = (1ll * cur3 * j) % M;
            dp[i] = (dp[i] + 1ll * cur4 * dp[i - j]) % M;
        }
    }
    printf("%dn", f[n]);
    return 0;
}

Problem solution in C.

#include<stdio.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 3001
typedef long long ll;
int f[N],g[N],c[N][N],n,p;
int main()
{
    scanf("%d%d",&n,&p);
    g[0]=1;
    f[1]=g[1]=1;
    f[2]=0; g[2]=1;
    fo(i,0,n) c[i][0]=1;
    fo(i,1,n)
        fo(j,1,i) c[i][j]=(c[i-1][j-1]+c[i-1][j])%p;
    f[0]=1;
    fo(i,3,n)
    {
        if (i&1)
        {
            f[i]=g[i-1]; 
            fo(j,i/2+1,i-1) f[i]=(f[i]-(ll)g[i-1-j]*f[j]%p*j%p*c[i-1][j])%p;
            int half=i/2;
            f[i]=(ll)f[i]*i%p; 
        } else
        {
            f[i]=g[i-1];
            fo(j,i/2,i-1)   f[i]=(f[i]-(ll)g[i-1-j]*f[j]%p*j%p*c[i-1][j])%p;
            f[i]=(ll)f[i]*i%p;
        }
        fo(j,1,i)
        g[i]=(g[i]+(ll)g[i-j]*c[i-1][j-1]%p*f[j]%p*j%p)%p;
    }
    printf("%dn",(f[n]+p)%p);
}