結果

問題 No.1944 ∞
ユーザー gew1fw
提出日時 2025-06-12 14:17:06
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,774 bytes
コンパイル時間 180 ms
コンパイル使用メモリ 83,032 KB
実行使用メモリ 124,544 KB
最終ジャッジ日時 2025-06-12 14:17:15
合計ジャッジ時間 4,685 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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ファイルパターン 結果
other AC * 13 WA * 2 TLE * 1 -- * 20
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ソースコード

diff # 

import math

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    N = int(data[idx])
    idx +=1
    X = int(data[idx])
    idx +=1
    Y = int(data[idx])
    idx +=1
    R = list(map(int, data[idx:idx+N]))
    idx += N
    
    sum_R = sum(R)
    D = math.hypot(X, Y)
    
    # Handle N=1 case
    if N == 1:
        if (X == 0 and Y == 0) or (X**2 + Y**2 == R[0]**2):
            print("Yes")
        else:
            print("No")
        return
    
    # For each possible R_last
    for r_last in R:
        # Compute the remaining elements
        temp = [x for x in R if x != r_last]
        if len(temp) == 0:
            # All elements are r_last, but since N >=2, len(temp) can't be zero here
            pass
        else:
            min_r_first = min(temp)
            max_r_first = max(temp)
            min_val = 2 * sum_R - max_r_first - r_last
            max_val = 2 * sum_R - min_r_first - r_last
            lower = D - r_last
            upper = D + r_last
            # Check overlap
            if (max_val >= lower) and (min_val <= upper):
                print("Yes")
                return
    # Check for the case where one R_last is the same as others, but minimal/maximal
    # Also, when R_last is the minimal or maximal, but R_first can be arranged
    # Additionally, check if the sum of all R's is sufficient
    # But for the third sample, this approach fails
    # So, perhaps another condition is needed
    # The problem is that the initial approach doesn't consider that the sum can be arranged to have |S| >= D - R_last even if S_total is larger
    # Thus, perhaps if sum_R >= D:
    if sum_R >= D:
        print("Yes")
        return
    print("No")

if __name__ == "__main__":
    main()
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