結果

問題 No.1417 100の倍数かつ正整数(2)
ユーザー brthyyjpbrthyyjp
提出日時 2021-03-05 23:11:05
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 678 ms / 3,000 ms
コード長 2,120 bytes
コンパイル時間 299 ms
コンパイル使用メモリ 82,368 KB
実行使用メモリ 183,860 KB
最終ジャッジ日時 2024-10-07 04:44:08
合計ジャッジ時間 8,847 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 * 3
other AC * 36
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 #
プレゼンテーションモードにする

s = str(input())
n = len(s)
#print(n)
mod = 10**9+7
dp = [[[0, 0] for _ in range(100)] for _ in range(n+1)]
dp[0][0][0] = 1
for i in range(n):
for j in range(100):
for k in range(2):
nd = int(s[i])
for d in range(10):
ni = i+1
if i == 0:
nj = d%100
else:
nj = (j*d)%100
nk = k
if k == 0:
if d > nd:
continue
elif d < nd:
nk = 1
else:
pass
if d != 0:
dp[ni][nj][nk] += dp[i][j][k]%mod
#print(dp)
#print((dp[n][0][0]+dp[n][0][1])%mod)
ans = (dp[n][0][0]+dp[n][0][1])%mod
N = 10**6
mod = 10**9+7
fac = [1]*(N+1)
finv = [1]*(N+1)
for i in range(N):
fac[i+1] = fac[i] * (i+1) % mod
finv[-1] = pow(fac[-1], mod-2, mod)
for i in reversed(range(N)):
finv[i] = finv[i+1] * (i+1) % mod
def cmb1(n, r, mod):
if r <0 or r > n:
return 0
r = min(r, n-r)
return fac[n] * finv[r] * finv[n-r] % mod
def prm1(n, r, mod):
if r <0 or r > n:
return 0
return fac[n] * finv[n-r] % mod
table9 = [0]*(10**5+50)
table9[0] = 1
for i in range(1, 10**5+50):
table9[i] = table9[i-1]*9
table9[i] %= mod
table4 = [0]*(10**5+50)
table4[0] = 1
for i in range(1, 10**5+50):
table4[i] = table4[i-1]*4
table4[i] %= mod
table5 = [0]*(10**5+50)
table5[0] = 1
for i in range(1, 10**5+50):
table5[i] = table5[i-1]*5
table5[i] %= mod
table8 = [0]*(10**5+50)
table8[0] = 1
for i in range(1, 10**5+50):
table8[i] = table8[i-1]*8
table8[i] %= mod
for i in range(n):
if i < 3:
continue
temp = table9[i]
# 5:0, 2, 6:0
temp -= table4[i]
# 5:0, 2, 6:1
temp -= i*2*table4[i-1]
# 5:1, 2, 6:0
temp -= i*1*table4[i-1]
# 5:1, 2, 6:1
temp -= cmb1(i, 2, mod)*2*2*table4[i-2]
#print(i, temp)
a = table9[i]-table8[i]-i*table8[i-1]
b = table9[i]-table5[i]-i*2*table5[i-1]
ans += a+b-temp
print(ans%mod)
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