QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#288663 | #6126. Sequence and Sequence | Crysfly | AC ✓ | 440ms | 8556kb | Python3 | 2.5kb | 2023-12-23 10:57:17 | 2023-12-23 10:57:17 |
Judging History
answer
from math import isqrt
# ======== 预处理:多项式 f 和 F 的点值表示 ========
# 多项式最大次数
MAX_K = 30
# prod[i][j] = product(i - k), 0 <= k <= j, k != i
prod = []
for i in range(MAX_K + 1):
prod.append([])
p = 1
for j in range(MAX_K + 1):
if i != j:
p *= i - j
prod[-1].append(p)
# 给 k 次多项式的点值表示 (0, y[0]), (1, y[1]), ..., (k, y[k])
# 利用拉格朗日插值法,求自变量为 x 时多项式的值
def eval(x, y):
if x >= 0 and x < len(y):
return y[x]
p = 1
for i in range(len(y)):
p *= x - i
nume, deno = 0, 1
for i in range(len(y)):
a = y[i] * (p // (x - i))
b = prod[i][len(y) - 1]
nume = nume * b + deno * a
deno *= b
return nume // deno
# 最大递归深度
MAX_DEP = 4
# polys[i] 是 f^{(i)}(x) 的点值表示
polys = []
polys.append([1])
# smPolys[i] 是 F^{(i)}(x) 的点值表示
smPolys = []
for _ in range(MAX_DEP):
prevLen = len(polys[-1])
# F^{(i)}(x) 是 f^{(i)}(x) 的前缀和
sm = [0]
for y in polys[-1][1:]:
sm.append(sm[-1] + y)
sm.append(sm[-1] + eval(prevLen, polys[-1]))
smPolys.append(sm)
# f^{(i + 1)}(x) = F^{(i)}(x * (x + 1) // 2 - 1)
sq = []
for x in range(0, prevLen * 2 + 1):
sq.append(eval(x * (x + 1) // 2 - 1, sm))
polys.append(sq)
# ======== 预处理:P 和 Q 小范围的值 ========
# 最大递归深度时,n < MAX_LEN
MAX_LEN = 610
P = [0]
t = 1
while len(P) <= MAX_LEN:
for _ in range(t + 1):
P.append(t)
t += 1
Q = [0, 1]
for i in range(2, MAX_LEN):
Q.append(Q[-1] + Q[P[i]])
# ======== 预处理:g(d, n) 在 d <= MAX_DEP,n < MAX_LEN 时的值 ========
# S[d][n] = g(d, n)
S = []
for i in range(MAX_DEP + 1):
S.append([0])
for j in range(1, MAX_LEN):
S[-1].append(S[-1][-1] + eval(j, polys[i]) * Q[P[j]])
# ======== 正式计算 ========
# 若 P(n) = p,则 (p + 3) * p // 2 >= n 且 p 最小
# 通过解二次方程算出来,因为开根号和除法都下取整了,要往后检查几个数
def calcP(n):
p = (isqrt(8 * n + 9) - 3) // 2
while p * (p + 3) // 2 < n:
p += 1
return p
def g(d, n):
if n < MAX_LEN:
return S[d][n]
np = calcP(n)
return eval(n, smPolys[d]) * g(0, np) - g(d + 1, np)
T = int(input())
for _ in range(T):
n = int(input())
print(g(0, n))
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 31ms
memory: 8556kb
input:
4 10 100 1000 987654321123456789
output:
30 2522 244274 235139898689017607381017686096176798
result:
ok 4 lines
Test #2:
score: 0
Accepted
time: 175ms
memory: 8428kb
input:
10000 613939207402503646 408978283345333197 976677512231308716 629053280913850466 148712339 236220313279945487 590396556995977994 9226 215693877607285701 649702683896705 343173887453826567 847003949499596615 867133040287550291 159928123569892354 864534948175618654 209739383170746721 4295456752378791...
output:
91030728117067063595428375419346402 40469246710473908695676971059074376 229951682342450470520349294486964970 95558039501640054006660579352994194 5418340556433412 13536357243714243974772906966693552 84197207203086091733385317631619504 20934656 11291075621624104092841708040651034 104019777231815308683...
result:
ok 10000 lines
Test #3:
score: 0
Accepted
time: 78ms
memory: 8496kb
input:
1000 6403632579734162001008235137760132245297 1307698664787972023762442022139627469666 668870338048562416595095770565441759482 5092270 8806864498496723812760785099973409980711 2178464116010775202899984038946879469187 204381824371 8638495456004772442511643693521120926431 45954412333082528168092594892...
output:
9822905445826021159291522774438593145331066315784567505849706373529921001845336 409552844852728078841401625660602682494169687360338453221088647649526339235330 107160056181509722327918304399871120167096186888511567354513496578559803370274 6354453295964 185817323129718525790559571287806776421589155460...
result:
ok 1000 lines
Test #4:
score: 0
Accepted
time: 115ms
memory: 8456kb
input:
2000 2587627816607030340103003175959756184662 7728753705064569253253827797978613582938 6507847628603052973674714721 6989857636824717968431061258300652290539 4734281027640913533293237760425416005062 9411123735455625690768098631226366597446 8309753930995253536640660321476246470149 63565157427098067709...
output:
1603610451790269237852635641930301658193367441164307312552842461667027137613454 14309943493171496191506053530749878345271155973702143153225815632926701434642842 10414697803791950572309383355053080338229465674803757518 11704102206894264239190418551798088411458157423624785417336589284698838535371266 5...
result:
ok 2000 lines
Test #5:
score: 0
Accepted
time: 235ms
memory: 8428kb
input:
5000 6701025283447923273597553918313900029215 1618190467906965189844764312914748628527 2135261797018456059451326428589353332126 8027429917075086154217136768450383650445 8263301530632040969183919589286944799002 376886964626613356031779878 1191561726595055348898524031899625958102 453561433135467095374...
output:
10756647971303093856509780939435968749671842310025383400207261624750784751725876 627115945498078452730193858129037650151913122482133071938783012929533045529940 1091921493223233296724616654299408127388059493196845968361252389122720100379070 154375707400033063668088306493199269136326791073281723548116...
result:
ok 5000 lines
Test #6:
score: 0
Accepted
time: 440ms
memory: 8496kb
input:
10000 260865660295317278841 5232352637620496342310336202478387251106 7108789244285764135987032973912918380 12766535319519586095540974948550152061 5138306300154916887884637635208203681949 7603163140266839847784708214115317398585 149590294591155205497765668766786424787 63283557694500045989299147454323...
output:
16323111740957704392106241109891718054228 6557703685144914472554701877112177422100848067214049191882271294010013621817762 12143115079716078114619105501427985631361994195400195527663921137836417758 39139456824156526604158618001888125076786817219954316014947704612553450312 6324051018379978443719363340...
result:
ok 10000 lines