結果
| 問題 |
No.1515 Making Many Multiples
|
| コンテスト | |
| ユーザー |
ansain
|
| 提出日時 | 2021-05-21 22:47:02 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 783 ms / 2,000 ms |
| コード長 | 2,250 bytes |
| コンパイル時間 | 216 ms |
| コンパイル使用メモリ | 82,300 KB |
| 実行使用メモリ | 108,460 KB |
| 最終ジャッジ日時 | 2024-10-10 09:29:26 |
| 合計ジャッジ時間 | 11,832 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 28 |
ソースコード
import sys
from collections import defaultdict, Counter, deque
from itertools import permutations, combinations, product, combinations_with_replacement, groupby, accumulate
import operator
from math import sqrt, gcd, factorial
# from math import isqrt, prod,comb # python3.8用(notpypy)
#from bisect import bisect_left,bisect_right
#from functools import lru_cache,reduce
#from heapq import heappush,heappop,heapify,heappushpop,heapreplace
#import numpy as np
#import networkx as nx
#from networkx.utils import UnionFind
#from numba import njit, b1, i1, i4, i8, f8
#from scipy.sparse import csr_matrix
#from scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra, bellman_ford, johnson, NegativeCycleError
# numba例 @njit(i1(i4[:], i8[:, :]),cache=True) 引数i4配列、i8 2次元配列,戻り値i1
def input(): return sys.stdin.readline().rstrip()
def divceil(n, k): return 1+(n-1)//k # n/kの切り上げを返す
def yn(hantei, yes='Yes', no='No'): print(yes if hantei else no)
def main():
mod = 10**9+7
mod2 = 998244353
n, k, x, y = map(int, input().split())
A = list(map(int, input().split()))
dp = [[-1]*k for i in range(k)]
dp[x % k][y % k] = 0
dp[y % k][x % k] = 0
max1 = [-1]*k
max2 = [-1]*k
max1[x % k] = 0
max1[y % k] = 0
max2[y % k] = 0
max2[x % k] = 0
for AA in A:
A2 = AA % k
for a in range(k):
b = (0-A2-a) % k
if dp[a][b] != -1:
dp[a][b] += 1
max1[a] = max(max1[a], dp[a][b])
max2[b] = max(max2[b], dp[a][b])
"""
if a != b and dp[b][a] != -1:
dp[b][a] += 1
max2[a] = max(max2[a], dp[b][a])
max1[b] = max(max1[b], dp[b][a])
"""
for c in range(k):
dp[c][A2] = max(dp[c][A2], max2[c], max1[c])
dp[A2][c] = max(dp[A2][c], max2[c], max1[c])
for c in range(k):
max1[c] = max(max1[c], dp[c][A2])
max2[A2] = max(max2[A2], dp[c][A2])
max2[c] = max(max2[c], dp[A2][c])
max1[A2]=max(max1[A2], dp[A2][c])
# print(dp,max1,max2)
print(max(max(b) for b in dp))
if __name__ == '__main__':
main()
ansain