結果

問題 No.243 出席番号(2)
ユーザー qwewe
提出日時 2025-05-14 13:05:06
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 18 ms / 2,000 ms
コード長 5,081 bytes
コンパイル時間 648 ms
コンパイル使用メモリ 74,612 KB
実行使用メモリ 6,272 KB
最終ジャッジ日時 2025-05-14 13:06:53
合計ジャッジ時間 1,782 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 30
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ソースコード

diff # 

#include <iostream>
#include <vector>
#include <numeric>

// Using long long for safety in calculations involving potentially large numbers
// Standard library includes useful functions and types.
using namespace std;

// Define the modulus as a constant integer.
// The problem requires calculations modulo 10^9 + 7.
const int MOD = 1000000007;

int main() {
    // Optimize standard input/output operations for speed.
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N; // Number of students
    cin >> N; // Read the number of students from input.

    // Vector `count` stores the frequency of each disliked number.
    // The size is N because attendance numbers are from 0 to N-1.
    // Disliked numbers A_i >= N are irrelevant because they can never be assigned.
    vector<int> count(N, 0); 
    for (int i = 0; i < N; ++i) {
        int A_i; // Disliked number for student S_i
        cin >> A_i; // Read the disliked number for student i.
        
        // We only care about constraints where the disliked number A_i is
        // within the range of possible attendance numbers [0, N-1].
        if (A_i >= 0 && A_i < N) {
            // Increment the count for the disliked number A_i.
            // This indicates one more student dislikes this number.
            count[A_i]++; 
        }
    }

    // Precompute factorials modulo MOD up to N.
    // `fact[k]` will store k! mod MOD. Factorials are needed for the inclusion-exclusion formula.
    vector<long long> fact(N + 1);
    fact[0] = 1; // Base case: 0! = 1
    for (int i = 1; i <= N; ++i) {
        // Calculate k! = (k-1)! * k.
        // Use `(long long)i` to cast `i` to long long before multiplication
        // to prevent potential overflow if `fact[i-1]` and `i` are large, even if their product fits in long long.
        fact[i] = (fact[i - 1] * (long long)i) % MOD; 
    }

    // Vector `C` will store the coefficients of the polynomial P(z) = product_{j=0}^{N-1} (1 + count[j] * z).
    // The coefficient C[k] (i.e., coefficient of z^k) represents the number of ways to choose k students
    // such that their disliked numbers are all distinct. This is the term denoted as C_k in the inclusion-exclusion formula.
    // The size is N+1 because the polynomial can have degree up to N.
    vector<long long> C(N + 1, 0);
    C[0] = 1; // Initialize the polynomial as P(z) = 1. The coefficient of z^0 is 1.
    int current_deg = 0; // Keep track of the current degree of the polynomial P(z).

    // Build the polynomial P(z) iteratively.
    // For each number j from 0 to N-1:
    for (int j = 0; j < N; ++j) {
        // If number j is disliked by at least one student (count[j] > 0):
        if (count[j] > 0) {
            long long c = count[j]; // c is the number of students who dislike number j.
            
            // Multiply the current polynomial (represented by coefficients in C) by the factor (1 + c*z).
            // Update the coefficients C[i] in place.
            // To correctly update in place, we must iterate from the highest possible new degree downwards.
            // This ensures that when calculating the new C[i], we use the value of C[i-1] from the *previous* polynomial state.
            // The formula for the new coefficient C'[i] is: C'[i] = old C[i] * 1 + old C[i-1] * c.
            for (int i = current_deg + 1; i >= 1; --i) {
                 // Calculate the term to add: (old C[i-1] * c) % MOD
                 long long term_to_add = (C[i - 1] * c) % MOD;
                 // Update C[i]: new C[i] = (old C[i] + term_to_add) % MOD
                 C[i] = (C[i] + term_to_add) % MOD;
            }
            // Note: C[0] remains unchanged because the update formula for i=0 would be C'[0] = old C[0] + old C[-1]*c. Since C[-1] is conceptually 0, C'[0]=old C[0].
            
            // After multiplying by (1 + c*z), the degree of the polynomial increases by 1.
            current_deg++;
        }
    }

    // Calculate the final answer using the principle of inclusion-exclusion.
    // The formula is: Answer = sum_{k=0}^{N} (-1)^k * C_k * (N-k)!
    long long ans = 0; // Initialize the answer to 0.
    
    // Iterate through k from 0 up to the computed degree of the polynomial P(z).
    // Note: C_k will be 0 for k > current_deg, so we don't need to sum further.
    for (int k = 0; k <= current_deg; ++k) {
        // Calculate the k-th term of the sum: C_k * (N-k)! mod MOD
        long long term = (C[k] * fact[N - k]) % MOD;
        
        // Add or subtract the term based on the sign (-1)^k.
        // Check if k is even or odd.
        if (k % 2 == 0) { // If k is even, the sign is +1.
            ans = (ans + term) % MOD; // Add the term to the answer.
        } else { // If k is odd, the sign is -1.
            // Subtract the term. Add MOD before taking modulo to handle potential negative results correctly.
            ans = (ans - term + MOD) % MOD; 
        }
    }

    // Output the final computed answer.
    cout << ans << endl;

    return 0; // Indicate successful execution.
}
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