結果

問題 No.3094 Stapler
コンテスト
ユーザー qwewe
提出日時 2025-05-14 13:06:44
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 5,141 bytes
コンパイル時間 391 ms
コンパイル使用メモリ 82,816 KB
実行使用メモリ 67,768 KB
最終ジャッジ日時 2025-10-23 22:37:52
合計ジャッジ時間 6,700 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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other RE * 72
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ソースコード

diff # 

import sys

# Function to read input and solve the problem
def solve():
    # Read the size of the character table N from standard input
    N = int(sys.stdin.readline())
    
    # Initialize an empty list to store the matrix A
    A = []
    # Initialize missing_pos to None. This will store the (row, column) index of the '?' entry.
    missing_pos = None
    
    # Read the N x N matrix from standard input, row by row
    for r in range(N):
        # Read a line, split it into strings based on whitespace
        line = sys.stdin.readline().split()
        # Initialize an empty list for the current row
        row = []
        for c in range(N):
            # Check if the current element string is '?'
            if line[c] == '?':
                # If it is '?', append None as a placeholder and store its 0-based position
                row.append(None) 
                missing_pos = (r, c)
            else:
                # If it's a number string, convert it to an integer and append to the row
                row.append(int(line[c]))
        # Append the completed row list to the matrix A
        A.append(row)

    # Handle the base case where N=1
    # The character table of the trivial group {e} is [[1]]. 
    # If N=1 and the input is '?', the missing value must be 1.
    if N == 1:
        print(1)
        return

    # Unpack the 0-based row and column index of the missing entry '?'
    r, c = missing_pos

    # Case 1: The missing entry is in the first column (c == 0).
    # This corresponds to an unknown character degree A[r][0].
    # We use the second orthogonality relation (column orthogonality).
    # For any two distinct columns j and k, sum_{i=0}^{N-1} chi_i(g_j) * conjugate(chi_i(g_k)) = 0.
    # Since character values are given as integers, the conjugate is the value itself.
    # So, sum_{i=0}^{N-1} A[i][j] * A[i][k] = 0 for j != k.
    # We apply this for columns j=0 and some k > 0.
    if c == 0: 
        # We need to find a column index k (1 <= k < N) such that A[r][k] is not zero.
        # A[r][k] is the value of character chi_r on the k-th conjugacy class representative.
        # Since the missing entry is A[r][0], all A[r][k] for k > 0 are known values.
        # It can be shown that for N > 1, such a k must exist for a valid character table structure.
        k = -1 # Initialize k to an invalid value
        for ki in range(1, N):
            # A[r][ki] is guaranteed to be a known integer because the missing element is at column 0.
            if A[r][ki] != 0:
                k = ki # Found a suitable column index k
                break
        
        # Calculate the sum S = sum_{i != r} A[i][0] * A[i][k].
        # This sum includes all terms in the orthogonality relation sum_{i=0}^{N-1} A[i][0] * A[i][k] = 0
        # except the term involving the missing value A[r][0].
        S = 0
        for i in range(N):
            if i != r: # Exclude the row r containing the missing element
                # Access known values A[i][0] and A[i][k]. These are known because the missing one is A[r][0].
                S += A[i][0] * A[i][k]
        
        # The full orthogonality equation is: S + A[r][0] * A[r][k] = 0.
        # Solve for the missing value A[r][0]: A[r][0] = -S / A[r][k].
        # A[r][k] is the known non-zero value we found.
        # Use integer division '//' since the problem guarantees an integer solution.
        missing_val = -S // A[r][k]
        print(missing_val)

    # Case 2: The missing entry is NOT in the first column (c > 0).
    else: 
        # We apply the second orthogonality relation between column c and the first column (k=0).
        # The relation states: sum_{i=0}^{N-1} A[i][c] * A[i][0] = 0 for c != 0.
        
        # The value A[r][0] is the degree of character chi_r. 
        # Since the missing value is A[r][c] with c > 0, A[r][0] is a known value.
        # Character degrees (first column entries) are positive integers, so A[r][0] >= 1, ensuring A[r][0] != 0.
        
        k = 0 # Use the first column (index 0) for orthogonality check.
        
        # Calculate the sum S = sum_{i != r} A[i][c] * A[i][k] = sum_{i != r} A[i][c] * A[i][0].
        # This sum includes all terms in the orthogonality relation sum_{i=0}^{N-1} A[i][c] * A[i][0] = 0
        # except the term involving the missing value A[r][c].
        S = 0
        for i in range(N):
            if i != r: # Exclude the row r containing the missing element
                # Access known values A[i][c] and A[i][0]. 
                # A[i][c] is known because the missing entry is A[r][c].
                # A[i][0] is known because the missing entry is not in column 0 (c > 0).
                 S += A[i][c] * A[i][0]
        
        # The full orthogonality equation is: S + A[r][c] * A[r][0] = 0.
        # Solve for the missing value A[r][c]: A[r][c] = -S / A[r][0].
        # A[r][0] is the known non-zero character degree.
        # Use integer division '//' as the result must be an integer.
        missing_val = -S // A[r][0]
        print(missing_val)

# Execute the solve function to run the program
solve()
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