Problem:


Score : 1100 points

Problem Statement

You are given integers N, K, and an integer sequence A of length M.

An integer sequence where each element is between 1 and K (inclusive) is said to be colorful when there exists a contiguous subsequence of length K of the sequence that contains one occurrence of each integer between 1 and K (inclusive).

For every colorful integer sequence of length N, count the number of the contiguous subsequences of that sequence which coincide with A, then find the sum of all the counts.
Here, the answer can be extremely large, so find the sum modulo 10^9+7.

Constraints

1 \leq N \leq 25000

1 \leq K \leq 400

1 \leq M \leq N

1 \leq A_i \leq K

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K M
A_1 A_2 ... A_M

Output

For every colorful integer sequence of length N, count the number of the contiguous subsequences of that sequence which coincide with A, then print the sum of all the counts modulo 10^9+7.

Sample Input 1

3 2 1
1

Sample Output 1

9

There are six colorful sequences of length 3: (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2) and (2,2,1).
The numbers of the contiguous subsequences of these sequences that coincide with A=(1) are 2, 2, 1, 2, 1 and 1, respectively.
Thus, the answer is their sum, 9.

Sample Input 2

4 2 2
1 2

Sample Output 2

12

Sample Input 3

7 4 5
1 2 3 1 2

Sample Output 3

17

Sample Input 4

5 4 3
1 1 1

Sample Output 4

0

Sample Input 5

10 3 5
1 1 2 3 3

Sample Output 5

1458

Sample Input 6

25000 400 4
3 7 31 127

Sample Output 6

923966268

Sample Input 7

9954 310 12
267 193 278 294 6 63 86 166 157 193 168 43

Sample Output 7

979180369



Sample input:
3 2 1
1


Sample output:
9

Submission status: Wrong Answer

Solution code:
MOD = 10 ** 9 + 7
n, k, m = map(int, input().split())
a = list(map(int, input().split()))
def include_perm(a):
    for i in range(m - k + 1):
        seen = [False] * (k + 1)
        for j in range(i, i + k):
            if seen[a[j]]:
                break
            seen[a[j]] = True
        else:
            return True
    return False
def unique_len(a):
    seen = [False] * (k+1)
    ans = 0
    for ai in a:
        if seen[ai]:
            break
        seen[ai] = True
        ans += 1
    return ans
if unique_len(a) < m:
    dp_fw = [[0] * k for _ in range(n-m+1)]
    dp_bw = [[0] * k for _ in range(n-m+1)]
    if not include_perm(a):
        dp_fw[0][unique_len(reversed(a))] = 1
        dp_bw[0][unique_len(a)] = 1
    for i in range(n-m):
        for j in range(k):
            dp_fw[i+1][j] = dp_fw[i][j]
            dp_bw[i+1][j] = dp_bw[i][j]
        for j in range(k-2, 0, -1):
            dp_fw[i+1][j] += dp_fw[i+1][j+1]
            dp_fw[i+1][j] %= MOD
            dp_bw[i+1][j] += dp_bw[i+1][j+1]
            dp_bw[i+1][j] %= MOD
        for j in range(k-1):
            dp_fw[i+1][j+1] += dp_fw[i][j] * (k-j)
            dp_fw[i+1][j] %= MOD
            dp_bw[i+1][j+1] += dp_bw[i][j] * (k-j)
            dp_bw[i+1][j] %= MOD
    ans = 0
    tot = pow(k, n-m, MOD)
    for i in range(n-m+1):
        lcnt = i
        rcnt = n - m - i
        t = tot - sum(dp_fw[rcnt]) * sum(dp_bw[lcnt])
        print(t)
        ans = (ans + t) % MOD
    print(ans)
else:
    dp = [[0] * k for _ in range(n+1)]
    dp2 = [[0] * k for _ in range(n+1)]
    dp[0][0] = 1
    for i in range(n):
        for j in range(1, k):
            dp[i+1][j] = dp[i][j]
            dp2[i+1][j] = dp2[i][j]
        for j in range(k-2, 0, -1):
            dp[i+1][j] += dp[i+1][j+1]
            dp[i+1][j] %= MOD
            dp2[i+1][j] += dp2[i+1][j+1]
            dp2[i+1][j] %= MOD
        for j in range(k-1):
            dp[i+1][j+1] += dp[i][j] * (k-j)
            dp[i+1][j+1] %= MOD
            dp2[i+1][j+1] += dp2[i][j] * (k-j)
            dp2[i+1][j+1] %= MOD
        for j in range(m, k):
            dp2[i+1][j] += dp[i+1][j]
            dp2[i+1][j] %= MOD
    f = 1
    for i in range(m):
        f = f * (k - i) % MOD
    t = sum(dp2[n]) * pow(f, MOD-2, MOD) % MOD
    ans = ((n-m+1) * pow(k, n-m, MOD) - t) % MOD
    print(ans)

Submission status: Accepted

Solution code:
import sys
import numpy as np
import numba
from numba import njit, b1, i4, i8, f8

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

MOD = 1_000_000_007

@njit((i8, i8), cache=True)
def mpow(a, n):
    p = 1
    while n:
        if n & 1:
            p = p * a % MOD
        a = a * a % MOD
        n >>= 1
    return p


@njit((i8, ), cache=True)
def fact_table(N):
    """Compute 0!, 1!, ..., (N-1)! and their inverses modulo MOD. 

    Parameters
    ----------
    N : int
        length of the output array. 

    Returns
    -------
    fact : np.ndarray
        0!, 1!, ..., (N-1)!

    fact_inv : np.ndarray
        inverse of 0!, 1!, ..., (N-1)! modulo MOD. 
    """
    fact = np.empty(N, np.int64)
    fact[0] = 1
    for n in range(1, N):
        fact[n] = n * fact[n - 1] % MOD
    fact_inv = np.empty(N, np.int64)
    fact_inv[N - 1] = mpow(fact[N - 1], MOD - 2)
    for n in range(N - 1, 0, -1):
        fact_inv[n - 1] = fact_inv[n] * n % MOD
    return fact, fact_inv

@njit((i8[:], i8), cache=True)
def is_colorful(A, K):
    if len(A) < K:
        return False
    counts = np.zeros(K + 1, np.int64)
    n_color = 0
    for i in range(K):
        x = A[i]
        if counts[x] == 0:
            n_color += 1
        counts[x] += 1
    if n_color == K:
        return True
    for i in range(K, len(A)):
        x = A[i]
        if counts[x] == 0:
            n_color += 1
        counts[x] += 1
        x = A[i - K]
        if counts[x] == 1:
            n_color -= 1
        counts[x] -= 1
        if n_color == K:
            return True
    return False

@njit((i8, i8, i8), cache=True)
def compute_dp(N, K, n):
    # 1,2,...,n に i 文字付け加えてできる非カラフル列
    # であって、suffix が j色のもの
    dp = np.zeros((N, K), np.int64)
    dp[0, n] = 1
    for i in range(1, N):
        # 色がかぶるときの遷移
        for k in range(K - 1, 0, -1):
            dp[i, k] += dp[i - 1, k]
            dp[i, k - 1] += dp[i, k]
        # 色が追加されるときの遷移
        for k in range(1, K):
            dp[i, k] += (K - k + 1) * dp[i - 1, k - 1]
        dp[i, 0] = 0
        dp[i] %= MOD
    return dp.sum(axis=1) % MOD

@njit((i8, i8, i8[:]), cache=True)
def main(N, K, A):
    M = len(A)
    total = (N - M + 1) * mpow(K, N - M) % MOD
    if is_colorful(A, K):
        return total

    # 非カラフル列の分を引きましょう
    def f(A):
        # 左から x 個までが相異なる
        ng = np.zeros(500, np.bool_)
        x = 0
        for a in A:
            if ng[a]:
                break
            ng[a] = 1
            x += 1
        return x

    l, r = f(A), f(A[::-1])

    if l < M:
        # 左右それぞれ独立の問題
        dpl = compute_dp(N, K, l)
        dpr = compute_dp(N, K, r)
        for i in range(N - M + 1):
            j = N - M - i
            total -= dpl[i] * dpr[j] % MOD
        return total % MOD

    # A がすべて異なる場合に解く
    # あらゆる長さ M の順列の登場回数を、順列の個数で割る

    # dp1 = 長さ、suffix の種類数 -> 個数
    dp1 = np.zeros((N + 1, K), np.int64)
    # dp2 = 長さ、suffix の種類数 -> 作った長さMの順列の個数
    dp2 = np.zeros((N + 1, K), np.int64)

    dp1[0, 0] = 1
    for i in range(1, N + 1):
        # 色がかぶるときの遷移
        for k in range(K - 1, 0, -1):
            dp1[i, k] += dp1[i - 1, k]
            dp1[i, k - 1] += dp1[i, k]
            dp2[i, k] += dp2[i - 1, k]
            dp2[i, k - 1] += dp2[i, k]
        # 色が追加されるときの遷移
        for k in range(1, K):
            dp1[i, k] += (K - k + 1) * dp1[i - 1, k - 1]
            dp2[i, k] += (K - k + 1) * dp2[i - 1, k - 1]
        dp1[i, 0] = 0
        dp1[i] %= MOD
        dp2[i, 0] = 0
        dp2[i] %= MOD
        # 新たに作った順列を加算
        for k in range(M, K):
            dp2[i, k] += dp1[i, k]
        dp2[i] %= MOD

    x = dp2[-1].sum() % MOD
    # これらの順列が A と一致する確率
    fact, fact_inv = fact_table(200_010)
    p = fact_inv[K] * fact[K - M] % MOD
    total -= x * p % MOD
    return total % MOD

N, K, M = map(int, readline().split())
A = np.array(read().split(), np.int64)

print(main(N, K, A))

Submission status: Accepted

Solution code:
MOD = 10 ** 9 + 7
n, k, m = map(int, input().split())
a = list(map(int, input().split()))
def include_perm(a):
    for i in range(m - k + 1):
        seen = [False] * (k + 1)
        for j in range(i, i + k):
            if seen[a[j]]:
                break
            seen[a[j]] = True
        else:
            return True
    return False
def unique_len(a):
    seen = [False] * (k+1)
    ans = 0
    for ai in a:
        if seen[ai]:
            break
        seen[ai] = True
        ans += 1
    return ans
if unique_len(a) < m:
    dp_fw = [[0] * k for _ in range(n-m+1)]
    dp_bw = [[0] * k for _ in range(n-m+1)]
    if not include_perm(a):
        dp_fw[0][unique_len(reversed(a))] = 1
        dp_bw[0][unique_len(a)] = 1
    for i in range(n-m):
        for j in range(k):
            dp_fw[i+1][j] = dp_fw[i][j]
            dp_bw[i+1][j] = dp_bw[i][j]
        for j in range(k-2, 0, -1):
            dp_fw[i+1][j] += dp_fw[i+1][j+1]
            dp_fw[i+1][j] %= MOD
            dp_bw[i+1][j] += dp_bw[i+1][j+1]
            dp_bw[i+1][j] %= MOD
        for j in range(k-1):
            dp_fw[i+1][j+1] += dp_fw[i][j] * (k-j)
            dp_fw[i+1][j] %= MOD
            dp_bw[i+1][j+1] += dp_bw[i][j] * (k-j)
            dp_bw[i+1][j] %= MOD
    ans = 0
    tot = pow(k, n-m, MOD)
    for i in range(n-m+1):
        lcnt = i
        rcnt = n - m - i
        t = tot - sum(dp_fw[rcnt]) * sum(dp_bw[lcnt])
        ans = (ans + t) % MOD
    print(ans)
else:
    dp = [[0] * k for _ in range(n+1)]
    dp2 = [[0] * k for _ in range(n+1)]
    dp[0][0] = 1
    for i in range(n):
        for j in range(1, k):
            dp[i+1][j] = dp[i][j]
            dp2[i+1][j] = dp2[i][j]
        for j in range(k-2, 0, -1):
            dp[i+1][j] += dp[i+1][j+1]
            dp[i+1][j] %= MOD
            dp2[i+1][j] += dp2[i+1][j+1]
            dp2[i+1][j] %= MOD
        for j in range(k-1):
            dp[i+1][j+1] += dp[i][j] * (k-j)
            dp[i+1][j+1] %= MOD
            dp2[i+1][j+1] += dp2[i][j] * (k-j)
            dp2[i+1][j+1] %= MOD
        for j in range(m, k):
            dp2[i+1][j] += dp[i+1][j]
            dp2[i+1][j] %= MOD
    f = 1
    for i in range(m):
        f = f * (k - i) % MOD
    t = sum(dp2[n]) * pow(f, MOD-2, MOD) % MOD
    ans = ((n-m+1) * pow(k, n-m, MOD) - t) % MOD
    print(ans)

Submission status: Runtime Error

Solution code:
import sys
import numpy as np
import numba
from numba import njit, b1, i4, i8, f8

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

MOD = 1_000_000_007

@njit((i8, i8), cache=True)
def mpow(a, n):
    p = 1
    while n:
        if n & 1:
            p = p * a % MOD
        a = a * a % MOD
        n >>= 1
    return p


@njit((i8, ), cache=True)
def fact_table(N):
    """Compute 0!, 1!, ..., (N-1)! and their inverses modulo MOD. 

    Parameters
    ----------
    N : int
        length of the output array. 

    Returns
    -------
    fact : np.ndarray
        0!, 1!, ..., (N-1)!

    fact_inv : np.ndarray
        inverse of 0!, 1!, ..., (N-1)! modulo MOD. 
    """
    fact = np.empty(N, np.int64)
    fact[0] = 1
    for n in range(1, N):
        fact[n] = n * fact[n - 1] % MOD
    fact_inv = np.empty(N, np.int64)
    fact_inv[N - 1] = mpow(fact[N - 1], MOD - 2)
    for n in range(N - 1, 0, -1):
        fact_inv[n - 1] = fact_inv[n] * n % MOD
    return fact, fact_inv

@njit((i8[:], i8), cache=True)
def is_colorful(A, K):
    if len(A) < K:
        return False
    counts = np.zeros(K + 1, np.int64)
    n_color = 0
    for i in range(K):
        x = A[i]
        if counts[x] == 0:
            n_color += 1
        counts[x] += 1
    if n_color == K:
        return True
    for i in range(K, len(A)):
        x = A[i]
        if counts[x] == 0:
            n_color += 1
        counts[x] += 1
        x = A[i - K]
        if counts[x] == 1:
            n_color -= 1
        counts[x] -= 1
        if n_color == K:
            return True
    return False

@njit((i8, i8, i8), cache=True)
def compute_dp(N, K, n):
    # 1,2,...,n に i 文字付け加えてできる非カラフル列
    # であって、suffix が j色のもの
    dp = np.zeros((N, K), np.int64)
    dp[0, n] = 1
    for i in range(1, N):
        # 色がかぶるときの遷移
        for k in range(K - 1, 0, -1):
            dp[i, k] += dp[i - 1, k]
            dp[i, k - 1] += dp[i, k]
        # 色が追加されるときの遷移
        for k in range(1, K):
            dp[i, k] += (K - k + 1) * dp[i - 1, k - 1]
        dp[i, 0] = 0
        dp[i] %= MOD
    return dp.sum(axis=1) % MOD

@njit((i8, i8, i8[:]), cache=True)
def main(N, K, A):
    M = len(A)
    total = (N - M + 1) * mpow(K, N - M) % MOD
    if is_colorful(A, K):
        return total

    # 非カラフル列の分を引きましょう
    def f(A):
        # 左から x 個までが相異なる
        ng = np.zeros(500, np.bool_)
        x = 0
        for a in A:
            if ng[a]:
                break
            ng[a] = 1
            x += 1
        return x

    l, r = f(A), f(A[::-1])

    if l < M:
        # 左右それぞれ独立の問題
        dp1 = compute_dp(N, K, l)
        dp2 = compute_dp(N, K, r)
        for i in range(N - M + 1):
            j = N - M - i
            total -= dp1[i] * dp2[j] % MOD
        return total % MOD

    # A がすべて異なる場合に解く
    # あらゆる長さ M の順列の登場回数を、順列の個数で割る

    # dp1 = 長さ、suffix の種類数 -> 個数
    dp1 = np.zeros((N + 1, K), np.int64)
    # dp2 = 長さ、suffix の種類数 -> 作った長さMの順列の個数
    dp2 = np.zeros((N + 1, K), np.int64)

    dp1[0, 0] = 1
    for i in range(1, N + 1):
        # 色がかぶるときの遷移
        for k in range(K - 1, 0, -1):
            dp1[i, k] += dp1[i - 1, k]
            dp1[i, k - 1] += dp1[i, k]
            dp2[i, k] += dp2[i - 1, k]
            dp2[i, k - 1] += dp2[i, k]
        # 色が追加されるときの遷移
        for k in range(1, K):
            dp1[i, k] += (K - k + 1) * dp1[i - 1, k - 1]
            dp2[i, k] += (K - k + 1) * dp2[i - 1, k - 1]
        dp1[i, 0] = 0
        dp1[i] %= MOD
        dp2[i, 0] = 0
        dp2[i] %= MOD
        # 新たに作った順列を加算
        for k in range(M, K):
            dp2[i, k] += dp1[i, k]
        dp2[i] %= MOD

    x = dp2[-1].sum() % MOD
    # これらの順列が A と一致する確率
    fact, fact_inv = fact_table(200_010)
    p = fact_inv[K] * fact[K - M] % MOD
    total -= x * p % MOD
    return total % MOD

def solve_naive(N, K, A):
    import itertools
    M = len(A)
    cnt = 0
    for p in itertools.product(range(1, K + 1), repeat=N):
        B = np.array(p).astype(np.int64)
        if is_colorful(B, K):
            for i in range(len(B) - M + 1):
                if np.all(B[i:i + M] == A):
                    cnt += 1
    return cnt

N, K, M = map(int, readline().split())
A = np.array(read().split(), np.int64)

print(main(N, K, A))