SRM 530

Div2 - 250 GogoXBallsAndBinsEasy
My intuition told me very simple solution. But I don't tried to prove it.

public class GogoXBallsAndBinsEasy {
    public int solve(int[] T) {
        int ans = 0;
        for (int i = 0; i < T.length / 2; ++i) {
            ans += T[T.length - 1 - i] - T[i];
        }
        return ans;
    }
}

Div2 - 500, Div1 - 250 GogoXCake

public class GogoXCake {
    public String solve(String[] cake, String[] cutter) {
        int n = cake.length;
        int m = cake[0].length();
        int r = cutter.length;
        int c = cutter[0].length();
        char a[][] = new char[n][m];
        for (int i = 0; i < n; ++i) {
            a[i] = cake[i].toCharArray();
        }
        for (int i = 0; i + r <= n; ++i) {
            for (int j = 0; j + c <= m; ++j) {
                boolean ok = true;
                for (int i1 = 0; i1 < r; ++i1) {
                    for (int j1 = 0; j1 < c; ++j1) {
                        if (cutter[i1].charAt(j1) != '.') {
                            continue;
                        }
                        if (a[i + i1][j + j1] != '.') {
                            ok = false;
                        }
                    }
                }
                if (!ok) {
                    continue;
                }
                for (int i1 = 0; i1 < r; ++i1) {
                    for (int j1 = 0; j1 < c; ++j1) {
                        if (cutter[i1].charAt(j1) == '.') {
                            a[i + i1][j + j1] = 'X';
                        }
                    }
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (a[i][j] == '.') {
                    return "NO";
                }
            }
        }
        return "YES";
    }
}

Div2 - 1000 GogoXReimuHakurai
Div1 - 500 GogoXMarisaKirisima
Despite the fact that these problems are different (second problem is more common) we can solve them using same approach.

public class GogoXMarisaKirisima {
    public int solve(String[] choices) {
        int n = choices.length;
        boolean a[][] = new boolean[n][n];
        for (int i = 0; i < n; ++i) {            
            for (int j = 0; j < n; ++j) {
                if (i == j) a[i][i] = true;
                else a[i][j] = choices[i].charAt(j) == 'Y';
            }
        }
        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (a[i][k] && a[k][j]) {
                        a[i][j] = true;
                    }
                }
            }
        }
        if (!a[0][n - 1]) return 0;
        int V = 0, E = 0;
        boolean use[] = new boolean[n];
        for (int i = 0; i < n; ++i) {
            use[i] = a[0][i] && a[i][n - 1];
        }
        for (int i = 0; i < n; ++i) {
            if (!use[i]) continue;
            ++V;
            for (int j = 0; j < n; ++j) {
                if (!use[j]) continue;
                if (choices[i].charAt(j) == 'Y') {
                    ++E;
                }
            }
        }
        return  E - V +  2;
    }
}

Div1 - 1000 GogoXBallsAndBins

import java.util.*;

public class GogoXBallsAndBins {
    int need, N;
    int[] T;
    Map m[][];
    static final long MOD = 1000 * 1000 * 1000 + 9;
    int rec(int k, int wait, int cum) {
        if (k == N) {
            if (cum == need && wait == 0) return 1;
            return 0;
        }
        if (m[k][wait].containsKey((Integer)cum)) 
            return (Integer)m[k][wait].get(cum);
        long res = rec(k + 1, wait, cum);
        long w = wait;
        if (wait > 0) {
            res += rec(k + 1, wait, cum) * w * 2; 
            res += rec(k + 1, wait - 1, cum + 2 * T[k]) * w * w;
        }
        res += rec(k + 1, wait + 1, cum - 2 * T[k]);
        res %= MOD;
        m[k][wait].put(cum, (int)res);
        return (int)res;
    }

    public int solve(int[] T, int moves) {
        this.T = T;
        need = 2 * moves;
        N = T.length;
        m = new Map[N + 1][N + 1];
        for (int i = 0; i <= N; ++i) {
            for (int j = 0; j <= N; ++j) {
                m[i][j] = new HashMap();
            }
        }
        return rec(0, 0, 0);
     }
}

SRM 527

Div 2 250 - P8XMatrixTransformation
It is obvious that we can transform source matrix to destination matrix if and only if number of ones in the first matrix equals number of ones in the second matrix (We always can swap any two elements of source matrix using given operation).

public class P8XMatrixTransformation {
    public String solve(String[] original, String[] target) {
        int n = original.length;
        int m = original[0].length();
        int c = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (original[i].charAt(j) == '1') ++c;
                if (target[i].charAt(j) == '1') --c;
            }
        }
        if (c == 0) return "YES";
        return "NO";
    }
}

Div2 550, Div1 275 - P8XGraphBuilder
As we can see from statement we have to build tree. Any tree has (N - 1) edges where N - number of vertexes. Since each edge (u, v) increase by one degree of u and v then sum of all vertexes degree in tree equals 2(E - 1) and each degree is not greater than N - 1 and not less than 1. If we become any array of degree (call it d) which satisfied previous restrictions we can always build tree using this array.  How we can do it? Sort array in non decreasing order and then let pick up first vertex (its degree is one) and connect it with first not one degree vertex with decreasing them degrees. Note that array keep non decreasing order. Repeat this operation until array would be as (1, 1) and add last edge. We become tree because in process we can't build cycle since we always added edges from left to right by array. And last action consists in finding optimal solution. We can do it using dynamic programming (like knapsack algorithm).

import java.util.Arrays;
public class P8XGraphBuilder {
    public int solve(int[] scores) {
        int N = scores.length + 1;
        int W = 2 * (N - 1);
        int dp[][] = new int[N + 1][W + 1];
        for (int i = 0; i <= N; ++i) {
            Arrays.fill(dp[i], Integer.MIN_VALUE);
        }
        dp[0][0] = 0;
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j < N; ++j) {
                int c = scores[j - 1];
                for (int k = j; k <= W; ++k) {
                    dp[i][k] = Math.max(dp[i][k], dp[i - 1][k - j] + c); 
                }
            }
        }
        return dp[N][W];
    }
}

Div1 1050 - P8XCoinChange

public class P8XCoinChange {
    static final int MOD = 1000 * 1000 + 3;

    int[][] mul(int[][] a, int[][] b) {
        int n = a.length;
        int c[][] = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                long cum = 0;
                for (int k = 0; k < n; ++k) {
                    cum += (long)a[i][k] * b[k][j];
                }
                c[i][j] = (int)(cum % MOD);
            }
        }
        return c;
    }
    int[][] pow(int a[][], long n) {
        int r[][] = new int[a.length][a.length];
        for (int i = 0; i < a.length; ++i) {
            r[i][i] = 1;
        }
        for (; n != 0; n >>= 1) {
            if ((n & 1) == 1) {
                r = mul(r, a);
            }
            a = mul(a, a);
        }
        return r;
    }
    public int solve(long sum, long[] vals) {
        int n = vals.length;
        int a[][][] = new int[n][n][n];
        for (int i = 0; i < n; ++i) {
            a[0][i][0] = 1;
        }
        for (int i = 1; i < n; ++i) {
            a[i] = pow(a[i - 1], vals[i] / vals[i - 1]);
            for (int k = i; k < n; ++k) {
                a[i][k][i]++;
            }
        }
        int r[][] = new int[n][n];
        for (int i = 0; i < n; ++i) {
            r[i][i] = 1;
        }
        for (int i = n - 1; i >= 0; --i) {
            long val = vals[i]; 
            long pw = sum / val;
            sum %= val;
            r = mul(r, pow(a[i], pw));
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += r[n - 1][i];
            ans %= MOD;
        }
        return ans;
     }
}

SRM 525

Div2 250 - RainyRoad
The problem has very short solution. Also it is possible solve it using more familiar approach (using dfs or bfs).

public class RainyRoad {
    public String isReachable(String[] road) {
        for (int i = 0; i < road[0].length(); ++i) {
            if (road[0].charAt(i) == 'W' && road[1].charAt(i) == 'W')
                return "NO";
        }
        return "YES";
    }
}

Div2 950 - MagicalSquare

I wrote solution with very big complexity O(N ^ 6). But it passed all tests.

public class MagicalSquare {
    String R[], C[];
    String A[][] = new String[3][3];
    Long dp[][][][] = new Long[3][51][51][51];

    long rec(int l, int i1, int i2, int i3) {
        if (l == 3) {
            if (i1 == C[0].length() && i2 == C[1].length() && i3 == C[2].length()) return 1;
            return 0;
        }
        if (dp[l][i1][i2][i3] != null) return dp[l][i1][i2][i3];
        String s1 = C[0].substring(i1);
        String s2 = C[1].substring(i2);
        String s3 = C[2].substring(i3);
        String cur = R[l];
        int len = cur.length();
        long res = 0;
        for (int i = 0; i <= len; ++i) {
            String t3 = cur.substring(i);
            if (!s3.startsWith(t3)) continue;
            for (int j = 0; j <= i; ++j) {
                String t1 = cur.substring(0, j);
                String t2 = cur.substring(j, i);
                if (!s2.startsWith(t2)) continue;
                if (!s1.startsWith(t1)) continue;
                res += rec(l + 1, i1 + t1.length(), i2 + t2.length(), i3 + t3.length());
            }
        }
        return dp[l][i1][i2][i3] = res;
    }

    public long getCount(String[] rowStrings, String[] columnStrings) {
        this.R = rowStrings;
        this.C = columnStrings;
        return rec(0, 0, 0, 0);
    }
}

Div1 300, Div2 600 - DropCoins

I wrote simple brute force solution with O(N^6) complexity which surprisingly do not became time limit. Of course there is more faster solution which uses inclusion–exclusion principle. It has O(N^4) complexity.

public class DropCoins {
    public int getMinimum(String[] board, int K) {
        int res = Integer.MAX_VALUE;
        int n = board.length;
        int m = board[0].length();
        for (int i1 = 0; i1 < n; ++i1)
        for (int i2 = i1; i2 < n; ++i2)
        for (int j1 = 0; j1 < m; ++j1)
        for (int j2 = j1; j2 < m; ++j2) {
            int cnt = 0;
            for (int i = i1; i <= i2; ++i)
                for (int j = j1; j <= j2; ++j)
                    if (board[i].charAt(j) != '.')
                        ++cnt;
            if (cnt != K) continue;
            int t1 = i1;
            int t2 = n - 1 - i2;
            int t = Math.min(2 * t1 + t2, 2 * t2 + t1);
            t1 = j1;
            t2 = m - 1 - j2;
            t += Math.min(2 * t1 + t2, 2 * t2 + t1);
            res = Math.min(res, t);
        }
        if (res == Integer.MAX_VALUE) return -1;
        return res;
    }
}

Div1 525 - Rumor

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

public class Rumor {

    int n, know, g[], mask;
    int res = Integer.MAX_VALUE;

    int check(int use) {
        int knowA = know, knowB = know;
        int giveA = 0, giveB = 0;
        int cnt = 0;
        while (knowA != mask || knowB != mask) {
            if (++cnt >= res) return -1;
            int newKnowA = 0, newKnowB = 0;
            for (int i = 0; i < n; ++i) {
                if (testBit(giveA, i) && testBit(giveB, i)) continue;
                if (!testBit(knowA, i) && !testBit(knowB, i)) continue;

                String s = "";
                if (!testBit(giveA, i) && testBit(knowA, i)) s += 'a';
                if (!testBit(giveB, i) && testBit(knowB, i)) s += 'b';
                if (s.length() == 0) continue;

                char ch;
                if (s.length() == 2) {
                    if (testBit(use, i)) ch = s.charAt(0);
                    else ch = s.charAt(1);
                } else ch = s.charAt(0);

                if (ch == 'a') {
                    newKnowA |= g[i];
                    giveA |= 1 << i;
                } else {
                    newKnowB |= g[i];
                    giveB |= 1 << i;
                }
            }
            if ((knowA | newKnowA) == knowA && (knowB | newKnowB) == knowB) return -1;
            knowA |= newKnowA;
            knowB |= newKnowB;
        }
        return cnt;
    }

    private boolean testBit(int use, int i) {
        return ((use >> i) & 1) == 1;
    }

    public int getMinimum(String knowledge, String[] graph) {
        n = knowledge.length();
        mask = (1 << n) - 1;
        g = new int[n];
        for (int i = 0; i < n; ++i) {
            if (knowledge.charAt(i) == 'Y') {
                know |= 1 << i;
            }
            for (int j = 0; j < n; ++j) {
                if (graph[i].charAt(j) == 'Y')
                    g[i] |= 1 << j;
            }
        }
        for (int use = 0; use < (1 << n); ++use) {
            int t = check(use);
            if (t == -1) continue;
            res = Math.min(res, t);
        }
        if (res == Integer.MAX_VALUE) return -1;
        return res;
    }
}

Div1 950 - MonochromePuzzle

public class MonochromePuzzle {
    public int getMinimum(String a[]) {
        int n = a.length;
        for (int i = 0; i < n; ++i) {
            int c = 0;
            for (int j = 0; j < n; ++j) {
                if (a[i].charAt(j) != a[j].charAt(i)) return -1;
                if (a[i].charAt(j) == '#') ++c;
            }
            if (c != 3) return -1;
        }
        int p[] = null;
        for (int i = 1; i < n; ++i) {
            p = check(a, 0, i);
            if (p != null) break;
        }
        if (p == null) return -1;


        int res = Integer.MAX_VALUE;

        for (int tt = 0; tt < 3; ++tt) {            
            for (int i = 0; i < n / 2; ++i) {
                int x = get(p);
                res = Math.min(x, res);
                rshift(p, 0, n / 2);
                lshift(p, n / 2, n);
            }
            reverse(p, 0, n);            

            for (int i = 0; i < n / 2; ++i) {
                int x = get(p);
                res = Math.min(x, res);
                rshift(p, 0, n / 2);
                lshift(p, n / 2, n);
            }

            reverse(p, 0, n / 2);
            reverse(p, n / 2, n);

            for (int i = 0; i < n / 2; ++i) {
                int x = get(p);
                res = Math.min(x, res);
                rshift(p, 0, n / 2);
                lshift(p, n / 2, n);
            }

            reverse(p, 0, n);
            for (int i = 0; i < n / 2; ++i) {
                int x = get(p);
                res = Math.min(x, res);
                rshift(p, 0, n / 2);
                lshift(p, n / 2, n);
            }


            if (n != 8) break;
            int np[] = new int[8];
            np[0] = p[0];
            np[1] = p[3];
            np[2] = p[4];
            np[3] = p[7];
            np[4] = p[6];
            np[6] = p[2];
            np[7] = p[1];
            np[5] = p[5];
            for (int i = 0; i < 8; ++i) p[i] = np[i];


        }
        return res;
    }

    private void reverse(int p[], int lo, int hi) {
        int n = (hi - lo) / 2;
        for (int i = 0; i < n; ++i) {
            int tmp = p[lo + i];
            p[lo + i] = p[hi - i - 1];
            p[hi - i - 1] = tmp;
        }
    }

    private void lshift(int p[], int lo, int hi) {
        int tmp = p[lo];
        for (int i = lo; i + 1 < hi; ++i) {
            p[i] = p[i + 1];
        }
        p[hi - 1] = tmp;
    }

    private void rshift(int p[], int lo, int hi) {
        int tmp = p[hi - 1];
        for (int i = hi - 1; i > 0; --i) {
            p[i] = p[i - 1];
        }
        p[0] = tmp;
    }

    private int get(int p[]) {
        int n = p.length;
        boolean use[] = new boolean[n];
        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (use[i]) continue;
            int cur = i;
            int c = 0;
            while (!use[cur]) {
                use[cur] = true;
                ++c;
                cur = p[cur];
            }
            if (c >= 2) res += c - 1;
        }
        return res;
    }

    private int[] check(String a[], int u, int v) {
        if (a[u].charAt(v) != '#') return null;
        int n = a.length;
        int p[] = new int[n];
        int left = 0, right = n - 1;
        p[left++] = u;
        p[right--] = v;
        boolean use[] = new boolean[n];
        use[u] = use[v] = true;
        int pu = u, pv = v;
        for (int k = 1; k < n / 2; ++k) {
            boolean ok = false;
            for (int i = 1; !ok && i < n; ++i) {
                if (use[i]) continue;
                for (int j = 1; !ok && j < n; ++j) {
                    if (use[j] || i == j) continue;
                    if (a[i].charAt(j) != '#') continue;
                    if (a[i].charAt(pu) != '#') continue;
                    if (a[j].charAt(pv) != '#') continue;
                    if (k == n / 2 - 1) {
                        if (a[i].charAt(u) != '#') continue;
                        if (a[j].charAt(v) != '#') continue;
                    }
                    ok = true;
                    p[left++] = i;
                    p[right--] = j;
                    use[i] = use[j] = true;
                    pu = i;
                    pv = j;
                }
            }
            if (!ok) return null;
        }
        return p;
    }
}

srm524

Problem set analysis

Div2 - 250 ShippingCubes

public class ShippingCubes {
 public int minimalCost(int N) {
  int r = Integer.MAX_VALUE;
  for (int i = 1; i <= N; ++i) {
   if (N % i != 0) continue;
   for (int j = 1; i * j <= N; ++j) {
    if (N % (i * j) != 0) continue;
    r = Math.min(r, i + j + N / i / j);
   }
  }
  return r;  
 }
}

Div2 - 500, Div1 - 250 Magic Diamond

public class MagicDiamonds {
 boolean isPrime(long n) {
  if (n < 2) 
   return false;
  for (long i = 2; i * i <= n; ++i) 
   if (n % i == 0) 
    return false;
  return true;
 }
 public long minimalTransfer(long n) {
  if (n == 3) return 3;
  if (!isPrime(n)) return 1;
  return 2;
 }
}

Div2 - 1000 MultiplesWithLimit

Let us build infinity tree in which from every node exist edges with weigth from set of available digits. Then every path from root to some node would build digit sequence, which matches some number X. Assign every node tree value X % N. Some nodes would be have same values. Let us start breath-first-search from every child root node and would be see nodes by asceding order edge weight. As soon as we visit node with value 0 stop search. Value X, which match this node is problem answer.

import java.util.*;
public class MultiplesWithLimit {
 String correct(String s) {
  int len = s.length();
  if (len < 9) return s;
  return s.substring(0, 3) + "..." + s.substring(len - 3) + "(" + len + " digits)";
 }
 public String minMultiples(int N, int[] fb) {
  Arrays.sort(fb);
  Queue<Integer> q = new LinkedList<Integer>();
  int d[] = new int[N + 1];
  int p[] = new int[N + 1];
  int pi[] = new int[N + 1];
  Arrays.fill(d, Integer.MAX_VALUE);
  Arrays.fill(p, -1);
  for (int i = 1; i < 10; ++i) {
   if (Arrays.binarySearch(fb, i) >= 0) continue;
   if (d[i % N] != Integer.MAX_VALUE) continue;
   q.add(i % N);
   d[i % N] = 1;
   pi[i % N] = i;
  }
  while (!q.isEmpty()) {
   int x = q.poll();
   if (x == 0) {
    String ans = "";
    for (; x != -1; x = p[x]) {
     ans = (char)('0' + pi[x]) + ans;
    }
    return correct(ans);
   }
   for (int i = 0; i < 10; ++i) {
    if (Arrays.binarySearch(fb, i) >= 0) continue;
    int nx = (x * 10 + i) % N;
    if (d[nx] == Integer.MAX_VALUE) {
     d[nx] = d[x] + 1;
     p[nx] = x;
     pi[nx] = i;
     q.add(nx);
    }
   }
  }
  return "IMPOSSIBLE";
 }
}

Div1 500 - LongestSequence
At first let's find out when we can build infinity sequence. Assume max is item from C array which has maximum absolute value. If we can build sequence from 2*max element, which satisfies given restrictions then obvious we can build sequence of length 3*max, 4*max an so on. So how we can check would we can build sequence of arbitrary length. Let us consider example C = {-2, 3}. Fix some sequence length and check it. For example fix length = 3. We want find out same array A of size 3 which satisfied given restriction. We can consider array of partials sums S, where S[i] = A[1] + A[2] + ... + A[i]. Its size equals size of array A. In this way sum A[i] + A[i + 1] + ... + A[j] equals S[j] - S[i - 1]. Thus we can take restrictions from array C and write out following equations:
S[2] - S[0] < 0
S[3] - S[1] < 0
S[3] - S[0] > 0
Transform these equations in such manner:
S[2] < S[0]
S[3] < S[1]
S[0] < S[3]
Consider graph which consists from 4 vertexes (from 0 to 3) and add edges which match these equations. Thus we add following edges: (2->0), (3->1), (0->3).
If we find out any cycle then we can't build sequence of length 3 because we become contradiction (if we consider length = 4, we are going to become a graph, which contains a cycle). Otherwise if we don't have a cycle then for prove that there is a suitable sequence of need length we can observe ideas from Cormen book (chapter "Difference contstraints and shortest pahts").
Now when we can check would we can build a sequence arbitrary length, we can with using binary search to find out problem answer.

import java.util.Arrays;

public class LongestSequence {
    int C[];
    boolean A[][] = new boolean[3000][3000];
    int Color[] = new int[3000];
    int N;

    void addEdge(int u, int v) {
        A[u][v] = true;
    }

    boolean dfs(int u) {
        Color[u] = 1;
        for (int v = 0; v <= N; ++v) {
            if (!A[u][v] || Color[v] == 2) continue;
            if (Color[v] == 1) return true;
            if (dfs(v)) return true;
        }
        Color[u] = 2;
        return false;
    }

    boolean check(int n) {
        N = n;
        Arrays.fill(Color, 0, n + 1, 0);
        for (int i = 0; i <= n; ++i) {
            Arrays.fill(A[i], 0, n + 1, false);
        }
        for (int x : C) {
            if (Math.abs(x) > n) continue;
            for (int i = Math.abs(x); i <= n; ++i) {
                if (x < 0) addEdge(i + x, i);
                if (x > 0) addEdge(i, i - x);
            }
        }
        for (int i = 0; i <= n; ++i) {
            if (Color[i] == 0 && dfs(i)) {
                return false;
            }
        }
        return true;
    }

    public int maxLength(int C[]) {
        this.C = C;
        if (check(2000)) return -1;
        int lo = 0, hi = 2000;
        while (hi - lo > 1) {
            int mid = hi + lo >> 1;
            if (check(mid)) lo = mid;
            else hi = mid;
        }
        return lo;

    }
}

Div1 - 1000 AxonometricProjection

import java.util.Arrays;

public class AxonometricProjection {

    static final int MOD = 1000 * 1000 * 1000 + 9;

    long P[] = new long[50 * 50 + 1];
    long P1[] = new long[50 * 50 + 1];
    int dp[][] = new int[100][100];
    int n2, m2, K;
    int C[][] = new int[51][51];
    {
        for (int i = 0; i < C.length; ++i) {
            C[i][0] = 1;
            for (int j = 1; j <= i; ++j) {
                C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
            }
        }
    }

    int rec(int n1, int m1) {
        if (n1 == 0 && m1 == 0) return 1;
        if (dp[n1][m1] != -1) return dp[n1][m1];
        long r = 0;
        for (int i = 0; i <= n1; ++i) {
            for (int j = 0; j <= m1; ++j) {
                if (i == n1 && j == m1) continue;
                long t1 = rec(i, j);
                long t2 = C[n1][i];
                long t3 = C[m1][j];
                int c1 = (n1 - i) * (m1 + m2) + (m1 - j) * (n1 + n2) - (n1 - i) * (m1 - j);
                long t4 = P[c1];
                r = (r + t1 * t2 % MOD * t3 % MOD * t4 % MOD) % MOD;
            }
        }
        int c = (n1 + n2) * (m1 + m2) - n2 * m2;
        r = (P1[c] - r) % MOD;
        if (r < 0) r += MOD;
        return dp[n1][m1] = (int)r;
    }

    public int howManyWays(int v[], int h[]) {
        Arrays.sort(v);
        Arrays.sort(h);
        if (v[v.length - 1] != h[h.length - 1]) return 0;
        long ans = 1;
        for (K = 1; K <= v[v.length - 1]; ++K) {
            int n1 = 0, n2 = 0;
            for (int cur : h) {
                if (cur == K) ++n1;
                if (cur > K) ++n2;
            }
            int m1 = 0, m2 = 0;
            for (int cur : v) {
                if (cur == K) ++m1;
                if (cur > K) ++m2;
            }
            if (m1 == 0 && n1 == 0) continue;
            P[0] = P1[0] = 1;
            int hi = (m1 + m2) * (n1 + n2);
            for (int i = 1; i <= hi; ++i) {
                P[i] = P[i - 1] * K % MOD;
                P1[i] = P1[i - 1] * (K + 1) % MOD;
            }
            this.n2 = n2;
            this.m2 = m2;
            for (int i = 0; i <= n1; ++i) {
                for (int j = 0; j <= m1; ++j) {
                    dp[i][j] = -1;
                }
            }
            ans = ans * rec(n1, m1) % MOD;
        }
        return (int)ans;
    }  
}