Overview

This problem asks us to find the number of ways to choose \(N\) distinct bento boxes from \(A \times B \times C\) types and assign them to \(N\) employees, modulo \(10^9 + 7\). This reduces to a permutation calculation.

Analysis

Rephrasing the Problem

A bento box is determined by a triple (main dish, side dish, dessert), and there are \(M = A \times B \times C\) types in total. The \(N\) employees are all distinct, and each must be assigned a different bento box.

This is equivalent to “choosing \(N\) items from \(M\) distinct items and distributing them to \(N\) people in order.” In other words, the answer is the permutation \(P(M, N)\).

\[P(M, N) = M \times (M-1) \times (M-2) \times \cdots \times (M-N+1)\]

Concrete Example

For example, when \(A = 2, B = 2, C = 2\), there are \(M = 8\) types of bento boxes. For \(N = 3\) people:

\[P(8, 3) = 8 \times 7 \times 6 = 336\]

• The 1st person has 8 choices, the 2nd person has 7 remaining choices, and the 3rd person has 6 remaining choices.

The Issue of \(M\) Being Huge

Since \(A, B, C\) can each be up to \(10^6\), \(M = A \times B \times C\) can be as large as \(10^{18}\). However, since we compute everything \(\mod (10^9 + 7)\), we don’t need to store the exact value of \(M\).

If we compute \(M \mod p\) (\(p = 10^9 + 7\)) first, each term \((M - i) \mod p\) can be obtained as follows:

\[(A \times B \times C - i) \mod p = ((A \times B \times C \mod p) - (i \mod p)) \mod p\]

This holds correctly due to the properties of modular arithmetic.

Algorithm

1. Compute \(M = A \times B \times C \mod (10^9 + 7)\)
2. Initialize \(\text{result} = 1\)
3. For \(i = 0, 1, 2, \ldots, N-1\), multiply \(\text{result}\) by \((M - i) \mod (10^9 + 7)\)
4. Output the final \(\text{result}\)

This directly computes \(P(M, N) \mod (10^9 + 7)\).

Complexity

• Time Complexity: \(O(N)\) — We simply loop \(N\) times performing multiplications (\(N \leq 10^6\), so this is sufficiently fast)
• Space Complexity: \(O(1)\) — Only a constant number of variables are used

Implementation Notes

• Preventing Overflow of \(M\): Since \(A \times B \times C\) can be as large as \(10^{18}\), we compute with \(\mod\) applied first. In Python, there is no overflow concern due to arbitrary-precision integers, but taking the modulus early makes the computation more efficient.

• Beware of Negative Remainders: \((M - i) \mod p\) can become negative (since \(M\) is the value after taking \(\mod\), when \(i\) is large, \(M - i\) can be negative). In Python, the % operator always returns a non-negative value, so no special handling is needed. However, in other languages (such as C++), you need to write ((M - i) % MOD + MOD) % MOD.

• Guarantee that \(N \leq A \times B \times C\): This is guaranteed by the problem statement, so as long as the values being multiplied don’t become \(0 \mod p\) along the way, the answer essentially won’t be \(0\) (though in the world of \(\mod p\) it may become \(0\), and that would simply be the correct answer).

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    N = int(input_data[0])
    A = int(input_data[1])
    B = int(input_data[2])
    C = int(input_data[3])
    
    MOD = 10**9 + 7
    
    # Total number of bento types: M = A * B * C
    # We need to assign N distinct bentos to N distinct people
    # This is P(M, N) = M * (M-1) * (M-2) * ... * (M-N+1)
    # where M = A * B * C
    
    # But M can be up to 10^18, so we can't compute M directly and then do modular arithmetic
    # Wait, actually we can. M mod p is fine, and then M-1 mod p, etc.
    # But M can be up to 10^18. We just need M mod MOD, (M-1) mod MOD, ..., (M-N+1) mod MOD
    
    # M = A * B * C mod MOD
    M = (A % MOD) * (B % MOD) % MOD * (C % MOD) % MOD
    
    # P(M, N) = M * (M-1) * ... * (M-N+1) mod MOD
    # We need to compute this product where each term is (M - i) mod MOD for i = 0, 1, ..., N-1
    # But M here is already reduced mod MOD. Since A*B*C can be up to 10^18,
    # we need to be careful: (M - i) mod MOD where M is the actual value A*B*C
    # Since we work mod MOD, (A*B*C - i) mod MOD = (M_mod - i % MOD) % MOD where M_mod = A*B*C mod MOD
    # This is correct because (A*B*C - i) mod MOD = ((A*B*C mod MOD) - (i mod MOD)) mod MOD
    
    result = 1
    for i in range(N):
        result = result * ((M - i) % MOD) % MOD
    
    print(result % MOD)

solve()

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This editorial was generated by claude4.6opus-thinking.