結果
| 問題 |
No.1417 100の倍数かつ正整数(2)
|
| コンテスト | |
| ユーザー |
ntuda
|
| 提出日時 | 2025-07-12 13:58:14 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,709 bytes |
| コンパイル時間 | 416 ms |
| コンパイル使用メモリ | 82,332 KB |
| 実行使用メモリ | 81,396 KB |
| 最終ジャッジ日時 | 2025-07-12 13:58:19 |
| 合計ジャッジ時間 | 4,757 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 25 WA * 11 |
ソースコード
'''
dp[i][j][k]:=
i桁目まで見て、
2の倍数がj (0..2)
5の倍数がk (0..2)の個数
'''
MOD = 10 ** 9 + 7
N = input()
LN = len(N)
dp = [[[0] * 3 for _ in range(3)] for _ in range(LN)]
dp[0][0][0] = 1
sum0 = 0
for i in range(LN - 1):
for j in range(1, 10):
jn = [0,0]
if j % 4 == 0:
jn = [2, 0]
elif j % 2 == 0:
jn = [1, 0]
elif j % 5 == 0:
jn = [0, 1]
for k in range(3):
for l in range(3):
k2 = min(2, k + jn[0])
l2 = min(2, l + jn[1])
dp[i + 1][k2][l2] += dp[i][k][l]
dp[i + 1][k2][l2] %= MOD
sum0 += dp[i + 1][2][2]
sum0 %= MOD
acn = [0, 0]
for i, n in enumerate(N):
n = int(n)
ni = LN - i - 1
Y = [[0] * 3 for _ in range(3)]
for j in range(3):
for k in range(3):
j2 = min(2, j + acn[0])
k2 = min(2, k + acn[1])
Y[j2][k2] += dp[ni][j][k]
Y[j2][k2] %= MOD
Z = [[0] * 3 for _ in range(3)]
dummy = 0
if i == LN - 1:
n = n + 1
for j in range(1, n):
jn = [0, 0]
if j % 4 == 0:
jn = [2, 0]
elif j % 2 == 0:
jn = [1, 0]
elif j % 5 == 0:
jn = [0, 1]
for k in range(3):
for l in range(3):
k2 = min(2, k + jn[0])
l2 = min(2, l + jn[1])
Z[k2][l2] += Y[k][l]
Z[k2][l2] %= MOD
sum0 += Z[2][2]
sum0 %= MOD
if n % 4 == 0:
acn[0] = min(2, acn[0] + 2)
elif n % 2 == 0:
acn[0] = min(2, acn[0] + 1)
elif n % 5 == 0:
acn[1] = min(2, acn[1] + 1)
print(sum0)
ntuda