「LOJ 138」类欧几里得算法

给出 $T$ 组询问,每组用 $n, a, b, c, k_1, k_2$ 来描述。对于每组询问,请你求出

$$
\sum_{x = 0} ^ {n} x ^ {k_1} {\left \lfloor \frac{ax + b}{c} \right \rfloor} ^ {k_2}
$$

对 $1000000007$ 取模。

链接

LOJ 138

题解

当 $a = 0$ 时,原式为
$$\lfloor \frac{b} {c} \rfloor ^ {k_2} \sum_{x = 0} ^ n x ^ {k_1}$$,此时就是一个自然数幂和,令 $B$ 表示伯努利数,$S_k(n) = \sum\limits_{x = 0} ^ n x ^ k$,则
$$S_k(n) = \frac {1} {k + 1}\sum_{j = 0} ^ k (-1) ^ j \binom{k + 1} {j} B_j n ^ {k + 1 - j}$$
令 $P_{i,j} = (-1) ^ j\binom{i + 1} {j} B_j$


当 $a \geq b$ 时,令 $a = qc + r, 0 \leq r \lt b$,原式为
$$\begin{aligned}\sum_{x = 0} ^ n x ^ {k_1} \lfloor \frac {ax + b} {c} \rfloor &= \sum_{x = 1} ^ n x ^ {k_1}(qx + \lfloor \frac{rx + b} {c} \rfloor) ^ {k_2} \\ &= \sum_{i = 0} ^ {k_2} \binom {k_2} {i}q ^ i \sum_{x = 0} ^ n x ^ {k_1 + i}\lfloor \frac{rx + b} {c} \rfloor ^ {k_2 - i}\end{aligned}$$。


当 $a \lt b$ 时,令 $y = \lfloor \frac{ax + b} {c} \rfloor$,则
$$\lfloor \frac{cy - b - 1} {a} \rfloor \lt x \leq \lfloor \frac{c(y + 1) - b - 1} {a} \rfloor $$
令 $m = \frac {an + b} {c}$,则
$$\sum_{x = 0} ^ n x ^ {k_1} \lfloor \frac{ax + b} {c} \rfloor ^ {k_2} = S_{k_1}(n)m ^ {k_2} + \sum_{y = 1} ^ m \big( (y - 1) ^ {k_2} - y ^ {k_2} \big)S_{k_1}(\lfloor \frac{cy - b - 1} {a} \rfloor)$$

$$(y - 1) ^ {k_2} - y ^ {k_2} = \sum_{i = 0} ^ {k_2 - 1}\binom{k_2} {i}(-1) ^ {k_2 - i}y ^ i$$

$$\sum_{x = 0} ^ {n}x ^ {k_1}\lfloor \frac {ax + b} {c} \rfloor ^ {k_2} = S_{k_1}(n)m ^ {k_2} + \sum_{i = 0} ^ {k_2 - 1}\sum_{h = 0} ^ {k_1 + 1}\big(\binom{k_2} {i}(-1) ^ {k_2 - i}P_{k_1, h}\sum_{y = 1} ^ m y ^ i\lfloor \frac{cy - b - 1} {c} \rfloor ^ h \big)$$

时间复杂度为 $O((k_1 + k_2) ^ 3\log(\min\{a, c\}))$。

代码

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/**
* Copyright (c) 2016-2018, xehoth
* All rights reserved.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
* http://www.apache.org/licenses/LICENSE-2.0
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
* 「LOJ 138」类欧几里得算法 18-05-2018
*
* @author xehoth
*/
#include <bits/stdc++.h>
const int MAXK = 10;
const int MOD = 1e9 + 7;
using uint = unsigned int;
using ull = unsigned long long;
inline int add(int x, int v) { return x + v >= MOD ? x + v - MOD : x + v; }
inline int dec(int x, int v) { return x - v < 0 ? x - v + MOD : x - v; }
int C[MAXK + 3][MAXK + 3], B[MAXK + 1], inv[MAXK + 3];
// P_{i,j} = (-1) ^ j\binom{i + 1} {j} B_j
int P[MAXK + 1][MAXK + 3];
inline int get(int k, int n) {
int ret = 0;
for (int i = 0; i < k + 2; i++) ret = ((ull)ret * n + P[k][i]) % MOD;
return (ull)ret * inv[k + 1] % MOD;
}
inline void init() {
inv[0] = inv[1] = 1;
for (int i = 2; i <= MAXK + 1; i++)
inv[i] = (MOD - MOD / i) * (ull)inv[MOD % i] % MOD;
for (int i = 0; i <= MAXK + 1; i++) {
C[i][0] = 1;
for (int j = 1; j <= i; j++)
C[i][j] = add(C[i - 1][j], C[i - 1][j - 1]);
}
B[0] = 1;
for (int i = 1, s; i <= MAXK; i++) {
s = 0;
for (int j = 0; j < i; j++) s = (s + (ull)C[i + 1][j] * B[j]) % MOD;
B[i] = (ull)(MOD - s) * inv[i + 1] % MOD;
}
for (int i = 0; i <= MAXK; i++) {
for (int j = 0; j <= i; j++) {
P[i][j] = (j & 1) ? ((ull)(MOD - C[i + 1][j]) * B[j] % MOD)
: ((ull)C[i + 1][j] * B[j] % MOD);
}
P[i][i + 1] = 0;
}
P[0][1] = 1;
}
/**
* \sum_{x = 0} ^ {n} x ^ {k_1} {\left \lfloor
* \frac{ax + b}{c} \right \rfloor} ^ {k_2}
*/
inline int classEuclid(int n, int a, int b, int c, int k1, int k2) {
using Info = std::tuple<int, int, int, int>;
static std::vector<Info> st;
st.clear();
st.reserve(std::__lg(std::min(a, c)) + 3);
for (;;) {
st.emplace_back(n, a, b, c);
if (n < 0 || a == 0) break;
if (a >= c) {
a %= c;
} else if (b >= c) {
b %= c;
} else {
n = ((long long)a * n + b) / c - 1;
b = c - b - 1;
std::swap(a, c);
}
}
const int S = k1 + k2;
static int buc[2][MAXK + 1][MAXK + 1];
using Ptr = int(*)[MAXK + 1];
Ptr f = buc[0], g = buc[1];
while (!st.empty()) {
std::tie(n, a, b, c) = st.back();
st.pop_back();
if (n < 0) {
continue;
} else if (a == 0) {
int q = b / c;
for (int i = 0, s; i <= S; i++) {
s = get(i, n);
for (int j = 0; j <= S - i; j++) {
g[i][j] = s;
s = (ull)s * q % MOD;
}
}
} else if (a >= c || b >= c) {
int q = (a >= c) ? a / c : b / c, d = (a >= c) ? 1 : 0;
// \sum_{i = 0} ^ {k_2} \binom {k_2} {i}q ^ i \sum_{x = 0} ^ n x ^
// {k_1 + i}\lfloor \frac{rx + b} {c} \rfloor ^ {k_2 - i}
for (int k1 = 0; k1 <= S; k1++) {
for (int k2 = 0, s; k2 <= S - k1; k2++) {
s = 0;
for (int i = 0, p = 1; i <= k2; i++) {
s = (s +
(ull)p * C[k2][i] % MOD * f[k1 + i * d][k2 - i]) %
MOD;
p = (ull)p * q % MOD;
}
g[k1][k2] = s % MOD;
}
}
} else {
// S_{k_1}(n)m ^ {k_2} + \sum_{i = 0} ^ {k_2 - 1}\sum_{h = 0} ^ {k_1
// + 1}\big(\binom{k_2} {i}(-1) ^ {k_2 - i}P_{k_1, h}\sum_{y = 1} ^
// m y ^ i\lfloor \frac{cy - b - 1} {c} \rfloor ^ h \big)
static int h[MAXK + 1][MAXK + 1];
for (int k2 = 0; k2 <= S - 1; k2++) {
for (int k1 = 0, s; k1 <= S - k2 - 1; k1++) {
s = 0;
for (int j = 0; j <= k1 + 1; ++j) {
s = (s + (ull)P[k1][k1 + 1 - j] * f[k2][j]) % MOD;
}
h[k1][k2] = (ull)s * inv[k1 + 1] % MOD;
}
}
const int m = ((ull)a * n + b) / c;
for (int k1 = 0; k1 <= S; k1++) {
int p = get(k1, n);
for (int k2 = 0, t; k2 <= S - k1; k2++) {
t = 0;
for (int i = 0; i < k2; i++)
t = (t + (ull)h[k1][i] * C[k2][i]) % MOD;
g[k1][k2] = dec(p, t);
p = (ull)p * m % MOD;
}
}
}
std::swap(f, g);
}
return f[k1][k2];
}
int main() {
// freopen("sample/1.in", "r", stdin);
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
int T, n, a, b, c, k1, k2;
init();
for (std::cin >> T; T--;) {
std::cin >> n >> a >> b >> c >> k1 >> k2;
std::cout << classEuclid(n, a, b, c, k1, k2) << '\n';
}
return 0;
}
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