結果

問題 No.823 Many Shifts Easy
ユーザー qwewe
提出日時 2025-05-14 13:26:29
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 62 ms / 2,000 ms
コード長 2,018 bytes
コンパイル時間 214 ms
コンパイル使用メモリ 82,256 KB
実行使用メモリ 59,904 KB
最終ジャッジ日時 2025-05-14 13:27:19
合計ジャッジ時間 1,406 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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ファイルパターン 結果
other AC * 10
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ソースコード

diff # 

import sys

def solve():
    N, K = map(int, sys.stdin.readline().split())

    MOD = 10**9 + 7

    if K == 0:
        # Initial state: pieces at 1, ..., N
        # Score = 1 + 2 + ... + N
        # N*(N+1) is always even.
        ans = (N * (N + 1) // 2) % MOD
        print(ans)
        return

    fact = [1] * (N + 1)
    inv_fact = [1] * (N + 1)
    for i in range(1, N + 1):
        fact[i] = (fact[i-1] * i) % MOD
    
    inv_fact[N] = pow(fact[N], MOD - 2, MOD)
    for i in range(N - 1, -1, -1): 
        inv_fact[i] = (inv_fact[i+1] * (i+1)) % MOD


    def nPk_mod(n_val, k_val):
        if k_val < 0 or k_val > n_val:
            return 0
        # n_val < 0 case should not occur with N >= 1
        return (fact[n_val] * inv_fact[n_val-k_val]) % MOD

    total_score_sum = 0
    
    # Term1 for C_j (j >= 1 and j not in A_S)
    # P(N-1, K)
    # This is non-zero if 0 <= K <= N-1
    term1_val = nPk_mod(N - 1, K)
    
    # Term2 for C_j (j in [1,N-1], j in A_S, j+1 in A_S, pos(j) < pos(j+1))
    # P(N-2, K-2) * (K choose 2)
    # This is non-zero if 2 <= K <= N
    term2_val = 0
    if K >= 2:
        # K*(K-1) is always even, so integer division //2 is fine before MOD.
        binom_K_2 = (K * (K - 1) // 2) 
        term2_val = (nPk_mod(N - 2, K - 2) * (binom_K_2 % MOD)) % MOD # binom_K_2 can be large
            
    # Cj_for_1_to_N_minus_1 is C_j for j in {1, ..., N-1}
    Cj_for_1_to_N_minus_1 = (term1_val + term2_val) % MOD
    
    # Sum for j from 1 to N-1
    # (sum of j from 1 to N-1) * Cj_for_1_to_N_minus_1
    sum_indices_1_to_N_minus_1 = 0
    if N > 1: # If N=1, sum is 0
        # (N-1)*N is always even
        sum_indices_1_to_N_minus_1 = ( (N - 1) * N // 2 ) % MOD 
    
    total_score_sum = (sum_indices_1_to_N_minus_1 * Cj_for_1_to_N_minus_1) % MOD
        
    # C_N (Count for N being occupied)
    # This is P(N-1, K), which is term1_val
    CN = term1_val 
    
    total_score_sum = (total_score_sum + N * CN) % MOD
        
    print(total_score_sum)

solve()
0