#593447 (PyPy3) No.1318 ABCD quadruplets

• 問題: No.1318 ABCD quadruplets
• ユーザー: chineristAC
• 提出日時: 2020-12-15 10:33:36
• 言語: PyPy3 (7.3.15)
• 結果: AC
• 実行時間: 1,136 ms / 2,000 ms
• コード長: 2,142 bytes
• コンパイル時間: 154 ms
• コンパイル使用メモリ: 82,244 KB
• 実行使用メモリ: 84,496 KB
• 最終ジャッジ日時: 2024-09-20 01:31:58
• 合計ジャッジ時間: 10,835 ms
• ジャッジサーバーID （参考情報）: judge5 / judge3

ファイルパターン	結果
sample	AC * 3
other	AC * 30

ソースコード

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N,M = map(int,input().split())
ans = [0 for i in range(N+1)]
cnt = 0
for a in range(M+1):
    for b in range(a+1,M+1):
        for c in range(b+1,M+1):
            for d in range(c+1,M+1):
                t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
                t //= 2
                if t<=N:
                    ans[t] += 24
                    cnt += 1
                else:
                    break
for a in range(M+1):
    b = a
    for c in range(a+1,M+1):
        for d in range(c+1,M+1):
            t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
            t //= 2
            if t<=N:
                ans[t] += 12
                cnt += 1
            else:
                break
for a in range(M+1):
    for b in range(a+1,M+1):
        c = b
        for d in range(b+1,M+1):
            t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
            t //= 2
            if t<=N:
                ans[t] += 12
                cnt += 1
            else:
                break
for a in range(M+1):
    for b in range(a+1,M+1):
        for c in range(b+1,M+1):
            d = c
            t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
            t //= 2
            if t<=N:
                ans[t] += 12
                cnt += 1
            else:
                break
for a in range(M+1):
    b = a
    for c in range(a+1,M+1):
        d = c
        t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
        t //= 2
        if t<=N:
            ans[t] += 6
            cnt += 1
        else:
            break
for a in range(M+1):
    b = a
    c = a
    for d in range(a+1,M+1):
        t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
        t //= 2
        if t<=N:
            ans[t] += 4
            cnt += 1
        else:
            break
for a in range(M+1):
    for b in range(a+1,M+1):
        c = b
        d = b
        t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
        t //= 2
        if t<=N:
            ans[t] += 4
            cnt += 1
        else:
            break
for a in range(M+1):
    b,c,d = a,a,a
    t = (a+b+c+d)**2+(a**2+b**2+c**2+d**2)
    t //= 2
    if t<=N:
        ans[t] += 1
        cnt += 1
    else:
        break
print(*ans,sep="\n")
 
 
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