結果

問題 No.990 N×Mマス計算(Kの倍数)
ユーザー ayaoni
提出日時 2020-11-24 18:26:08
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,242 bytes
コンパイル時間 427 ms
コンパイル使用メモリ 82,416 KB
実行使用メモリ 111,216 KB
最終ジャッジ日時 2024-07-23 18:43:53
合計ジャッジ時間 2,965 ms
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 11 WA * 8
権限があれば一括ダウンロードができます

ソースコード

diff #

import sys
from math import gcd
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LS(): return list(sys.stdin.readline().rstrip().split())


N,M,K = MI()
X = LS()
op = X[0]
B = [int(X[i]) for i in range(1,M+1)]
A = [I() for _ in range(N)]

if op == '+':
    count_A = {}
    count_B = {}
    for i in range(N):
        a = A[i] % K
        if a in count_A.keys():
            count_A[a] += 1
        else:
            count_A[a] = 1
    for i in range(M):
        b = B[i] % K
        if b in count_B.keys():
            count_B[b] += 1
        else:
            count_B[b] = 1
    ans = 0
    for i in count_A.keys():
        j = (K-i) % K
        if j in count_B.keys():
            ans += count_A[i]*count_B[j]
else:
    count_A = {}
    count_B = {}
    for a in A:
        g = gcd(a,K)
        if g in count_A.keys():
            count_A[g] += 1
        else:
            count_A[g] = 1
    for b in A:
        g = gcd(b,K)
        if g in count_B.keys():
            count_B[g] += 1
        else:
            count_B[g] = 1
    ans = 0
    for d in count_A.keys():
        e = K//d
        if e in count_B.keys():
            ans += count_A[d]*count_B[e]

print(ans)
0