結果

問題 No.398 ハーフパイプ(2)
コンテスト
ユーザー lam6er
提出日時 2025-03-20 20:52:41
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 147 ms / 2,000 ms
コード長 3,242 bytes
コンパイル時間 180 ms
コンパイル使用メモリ 81,648 KB
実行使用メモリ 77,668 KB
最終ジャッジ日時 2025-03-20 20:52:56
合計ジャッジ時間 3,063 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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ファイルパターン 結果
other AC * 17
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ソースコード

diff # 

import math
from math import comb

def main():
    x_str = input().strip()
    
    # Split the input into integral and fractional parts
    if '.' in x_str:
        integral_part, fractional_part = x_str.split('.')
        fractional_part = fractional_part.ljust(2, '0')[:2]
    else:
        integral_part = x_str
        fractional_part = '00'
    
    integral = int(integral_part)
    fractional = int(fractional_part[:2])
    
    # Calculate S = 4*X as an integer
    # X is integral_part + fractional_part/100
    # So 4*X = 4*integral_part + (4*fractional_part)/100
    # To avoid floating points, we compute S as (4 * (integral * 100 + fractional)) // 100
    total = integral * 100 + fractional
    S = (4 * total) // 100
    
    ans = 0
    
    # Case 1: All scores are the same (L == H)
    if fractional == 0 and 0 <= integral <= 100:
        ans += 1  # All six judges give X
    
    # Case 2: L < H
    for L in range(0, 100):
        for H in range(L + 1, 101):
            # Iterate over possible a and b where a >=1, b >=1, a + b <=6
            for a in range(1, 6):
                max_b = 6 - a
                if max_b < 1:
                    continue
                for b in range(1, max_b + 1):
                    k = 6 - a - b
                    sum_m = S + L + H - L * a - H * b
                    
                    if k == 0:
                        # Check sum_m is 0 and equation (a-1)*L + (5-a)*H == S
                        if sum_m == 0 and (4 * total) // 100 == (a-1)*L + (5 - a)*H:
                            # Compute combinations: C(6, a) * C(6-a, b)
                            c = comb(6, a) * comb(6 - a, b)
                            ans += c
                    else:
                        # Check H - L >= 2, sum_m is within the valid range
                        if H - L < 2:
                            continue
                        
                        lower = k * (L + 1)
                        upper = k * (H - 1)
                        
                        if sum_m < lower or sum_m > upper:
                            continue
                        
                        # Calculate the number of ways for the intermediate values
                        D = H - L - 2
                        Y = sum_m - k * (L + 1)
                        
                        if Y < 0 or Y > k * D:
                            continue
                        
                        # Apply inclusion-exclusion principle
                        max_t = Y // (D + 1)
                        current = 0
                        for t in range(0, max_t + 1):
                            sign = (-1) ** t
                            term = comb(k, t)
                            rem = Y - t * (D + 1)
                            term *= comb(rem + k - 1, k - 1)
                            current += sign * term
                        
                        if current <= 0:
                            continue
                        
                        # Add to answer considering all arrangements
                        c_ab = comb(6, a) * comb(6 - a, b)
                        ans += c_ab * current
    print(ans)

if __name__ == "__main__":
    main()
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