「BZOJ 3453」XLkxc-拉格朗日插值


$$\sum_{i=0}^{n} \sum_{j=1}^{a+i \times d} \sum_{l=1}^{j}l^k$$

链接

BZOJ 3453

题解

令 $f(n) = \sum\limits_{i = 1} ^ n i ^ k$,则 $f$ 为一个 $k$ 次多项式的前缀和,故 $f$ 为 $k + 1$ 次多项式。
令 $g(n) = \sum\limits_{i = 1} ^ n f(i)$,$g$ 为 $k + 2$ 次多项式。
令 $h(x) = g(a + x \times d)$,为 $k + 2$ 次多项式。
所以我们可以用快速幂在 $O(k \log k)$ 的时间内预处理 $g$ 的前 $k + 3$ 项,然后用拉格朗日插值在 $O(k)$ 的时间内得到一个 $h(x)$,$O(k ^ 2)$ 插出 $k + 4$ 个 $h$ 并求前缀和得到 $k + 4$ 个 $ans(x)$,再对其插值得到 $ans(n)$。

时间复杂度 $O(k ^ 2)$。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 3453」XLkxc 18-09-2017
* 拉格朗日插值
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig ? x = -x : 0;
}
inline void read(char &c) {
while (c = read(), isspace(c) && c != -1)
;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void print(const char *s) {
for (; *s; s++) print(*s);
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
struct InputOutputStream {
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
~InputOutputStream() { flush(); }
} io;
}
namespace {
using IO::io;
typedef unsigned int uint;
const uint MOD = 1234567891;
const int MAXN = 123;
typedef unsigned long long ulong;
inline int modPow(int a, int b) {
register int ret = 1;
for (; b; b >>= 1, a = (ulong)a * a % MOD)
(b & 1) ? ret = (ulong)ret * a % MOD : 0;
return ret;
}
struct Task {
uint inv[MAXN + 10];
int k, a, n, d;
inline void init(const int n) {
inv[0] = inv[1] = 1;
for (register int i = 2; i <= n; i++)
inv[i] = (ulong)i * inv[i - 1] % MOD;
inv[n] = modPow(inv[n], MOD - 2);
for (register int i = n - 1; i >= 0; i--)
inv[i] = inv[i + 1] * (i + 1ll) % MOD;
}
inline uint interpolation(uint *f, uint u, int n) {
static uint pre[MAXN + 10], suf[MAXN + 10];
pre[0] = suf[n + 2] = 1;
for (register int i = 1; i <= n + 1; i++)
pre[i] = (ulong)pre[i - 1] * (u - i + MOD) % MOD;
for (register int i = n + 1; i; i--)
suf[i] = (ulong)suf[i + 1] * (u - i + MOD) % MOD;
register uint ret = 0, tmp;
for (register int i = 1; i <= n + 1; i++) {
tmp = (ulong)f[i] * pre[i - 1] % MOD * suf[i + 1] % MOD *
inv[i - 1] % MOD * inv[n + 1 - i] % MOD;
if ((n + 1 - i) % 2) tmp = MOD - tmp;
ret = (ret + tmp) % MOD;
}
return ret;
}
inline void solve() {
register int T;
io >> T;
init(MAXN + 5);
static uint g[MAXN + 10], f[MAXN + 10];
while (T--) {
io >> k >> a >> n >> d;
for (register int i = 0; i <= k + 3; i++) g[i] = modPow(i, k);
for (register int i = 1; i <= k + 3; i++)
g[i] = (g[i] + g[i - 1]) % MOD;
for (register int i = 1; i <= k + 3; i++)
g[i] = (g[i] + g[i - 1]) % MOD;
f[0] = interpolation(g, a, k + 2);
for (register int i = 1; i <= k + 5; i++)
f[i] = (f[i - 1] +
interpolation(g, (a + (ulong)d * i) % MOD, k + 2)) %
MOD;
io << interpolation(f, n, k + 4) << '\n';
}
}
} task;
}
int main() {
task.solve();
return 0;
}
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