結果
| 問題 |
No.42 貯金箱の溜息
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 00:14:06 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 111 ms / 5,000 ms |
| コード長 | 2,909 bytes |
| コンパイル時間 | 163 ms |
| コンパイル使用メモリ | 81,432 KB |
| 実行使用メモリ | 77,360 KB |
| 最終ジャッジ日時 | 2025-04-16 00:15:23 |
| 合計ジャッジ時間 | 946 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 |
|---|---|
| other | AC * 3 |
ソースコード
MOD = 10**9 + 9
# Precompute C(k) for k from 0 to 599
max_k = 599
C = [0] * (max_k + 1)
for k in range(max_k + 1):
cnt = 0
max_e = k // 100
for e in range(max_e + 1):
rem_k = k - 100 * e
max_d = rem_k // 20
for d in range(max_d + 1):
rem_k2 = rem_k - 20 * d
max_c = rem_k2 // 10
for c in range(max_c + 1):
rem_k3 = rem_k2 - 10 * c
max_b = rem_k3 // 2
cnt += max_b + 1
C[k] = cnt % MOD
# Compute ans(q) as the prefix sum of C
ans = [0] * (max_k + 1)
current = 0
for k in range(max_k + 1):
current = (current + C[k]) % MOD
ans[k] = current
# For each residue r (0..99), collect the values ans(100m + r) for m=0,1,2,3,4,5
# Then perform Lagrange interpolation to find the coefficients of the polynomial
# Since the polynomial is of degree 5, we need 6 points
# We'll store for each residue r the coefficients of the polynomial a_5 m^5 + ... + a_0
from itertools import product
def lagrange_interpolation(xs, ys, mod):
n = len(xs)
res = [0] * n
for i in range(n):
numerator = [1]
for j in range(n):
if i == j:
continue
numerator = mul_poly(numerator, [-xs[j], 1], mod)
denom = 1
for j in range(n):
if i == j:
continue
denom = denom * (xs[i] - xs[j]) % mod
inv_denom = pow(denom, mod-2, mod)
numerator = [y * inv_denom % mod for y in numerator]
numerator = [y * ys[i] % mod for y in numerator]
res = add_poly(res, numerator, mod)
return res
def mul_poly(a, b, mod):
res = [0] * (len(a) + len(b) - 1)
for i in range(len(a)):
for j in range(len(b)):
res[i+j] = (res[i+j] + a[i] * b[j]) % mod
return res
def add_poly(a, b, mod):
res = [0] * max(len(a), len(b))
for i in range(len(a)):
res[i] = (res[i] + a[i]) % mod
for i in range(len(b)):
res[i] = (res[i] + b[i]) % mod
return res
# Precompute coefficients for each residue
coeff = {}
for r in range(100):
xs = []
ys = []
for m in range(6):
q = 100 * m + r
if q > max_k:
break
xs.append(m)
ys.append(ans[q])
if len(xs) < 6:
# Not enough points, but in our case, max_k=599, r=99: 100*5+99=599, so all residues have 6 points
pass
poly = lagrange_interpolation(xs, ys, MOD)
coeff[r] = poly
# Read input and process each test case
import sys
input = sys.stdin.read().split()
T = int(input[0])
cases = list(map(int, input[1:T+1]))
for M in cases:
q = M // 5
r = q % 100
m = (q - r) // 100
if m < 0:
print(0)
continue
poly = coeff[r]
res = 0
for i, c in enumerate(poly):
term = c
term = term * pow(m, i, MOD) % MOD
res = (res + term) % MOD
print(res)
lam6er