結果

問題 No.3072 Speedrun Query
ユーザー qwewe
提出日時 2025-05-14 12:55:54
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 3,059 bytes
コンパイル時間 386 ms
コンパイル使用メモリ 82,664 KB
実行使用メモリ 80,876 KB
最終ジャッジ日時 2025-05-14 12:57:42
合計ジャッジ時間 7,377 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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ファイルパターン 結果
sample RE * 1
other RE * 21
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ソースコード

diff # 

# -*- coding: utf-8 -*-
import sys
import decimal

# Set the precision for decimal module calculations.
# The problem requires a very high precision (absolute or relative error <= 10^-15).
# Standard floats (double precision) might not be sufficient, especially for
# square roots of large numbers close to perfect squares, or when summing many numbers.
# The maximum possible sum can be around N * sqrt(10^18) = 10^6 * 10^9 = 10^15.
# To maintain precision when adding potentially small sqrt values to a large running sum,
# and to ensure the final result meets the 1e-15 tolerance, we need sufficient
# working precision. A precision of 40 decimal digits provides a good safety margin.
decimal.getcontext().prec = 40

# Read the number of elements in the sequence
N = int(sys.stdin.readline())

# Initialize the running sum as a Decimal object with value 0
current_sum = decimal.Decimal(0)

# Use optimized I/O functions for potentially large input/output
# sys.stdin.readline is generally faster than input()
# sys.stdout.write is generally faster than print()
readline = sys.stdin.readline
write = sys.stdout.write

# Process each number in the sequence
for _ in range(N):
    # Read the next number as a string and remove leading/trailing whitespace
    x_str = readline().strip()

    # Convert the string representation to a Decimal object.
    # We assume the input consists of valid non-negative integers as per constraints.
    x_dec = decimal.Decimal(x_str)

    # Calculate the square root only if the number is positive.
    # The square root of 0 is 0, and adding 0 does not change the sum.
    # The problem constraints state x_i >= 0.
    if x_dec > 0:
        # Compute the square root using Decimal's arbitrary precision sqrt method.
        # The calculation uses the precision set in the context (40 digits).
        sqrt_x = x_dec.sqrt()
        # Add the computed square root to the running sum.
        # Decimal addition maintains the precision.
        current_sum += sqrt_x

    # Format the current cumulative sum for output.
    # The problem requires high precision, but also has an output size limit (~20MB).
    # Printing with a fixed high number of decimal places (like %.17f) might exceed the size limit.
    # The format specifier "{:.17g}" prints the number using up to 17 significant digits.
    # It uses fixed-point notation for numbers where it's suitable, and scientific
    # notation for very large or very small numbers, potentially reducing output length.
    # This format is chosen as a balance between satisfying the 1e-15 precision requirement
    # (often met via the relative error condition for large results) and managing output size.
    output_str = f"{current_sum:.17g}"

    # Write the formatted result to standard output, followed by a newline character.
    write(output_str + '\n')

# Flushing stdout might be necessary in some environments, but typically not required
# in competitive programming platforms as the buffer flushes upon program termination
# or when it fills up.
# sys.stdout.flush()
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