Luogu3338 [ZJOI2014]力
作者:互联网
Description
给定一个长度为\(n\)的数列\(q_i\),定义
\[
F_j = \sum_{i < j} \frac{q_i q_j}{(i - j)^2} - \sum_{i > j} \frac{q_i q_j}{(i - j)^2} \\
E_i = \frac{F_i}{q_i}
\]
求出所有的\(E_i\)
Solution
直接上推导过程
\[
E_j = \frac{\sum_{i < j} \frac{q_i q_j}{(i - j)^2} - \sum_{i > j} \frac{q_i q_j}{(i - j)^2}}{q_j} \\
= \sum_{i < j} \frac{q_i}{(i - j)^2} - \sum_{i > j} \frac{q_i}{(i - j)^2}
\]
于是我们可以令
\[
F(x) = q_x \\
G(x) =
\begin{cases}
-\frac{1}{x^2}~~~(x < 0) \\
0~~~~~~~~(x = 0) \\
\frac{1}{x^2}~~~~~~(x > 0)
\end{cases}
\]
则(下标默认从0开始)
\[
E_j = \sum_{i = 0}^{n - 1} F(j) \cdot G(j - i)
\]
所以这显然是一个卷积形式,所以我们就直接FFT求出卷积的系数即\(E_i\)即可
Code
#include <bits/stdc++.h>
using namespace std;
#define fst first
#define snd second
#define mp make_pair
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef pair<int, int> pii;
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
inline int read() {
int sum = 0, fg = 1; char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
return fg * sum;
}
namespace FFT {
const int MAX_LEN = 1 << 19;
const double PI = acos(-1.0);
struct com {
double a, b;
com (double _a = 0.0, double _b = 0.0): a(_a), b(_b) { }
com operator + (const com &t) const { return com(a + t.a, b + t.b); }
com operator - (const com &t) const { return com(a - t.a, b - t.b); }
com operator * (const com &t) const { return com(a * t.a - b * t.b, a * t.b + b * t.a); }
};
int cnt, len, rev[MAX_LEN];
com g[MAX_LEN];
void init(int N) {
for (cnt = -1, len = 1; len <= N; len <<= 1) ++cnt;
for (int i = 0; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << cnt);
g[0] = com(1.0, 0.0);
com G(cos(PI * 2 / len), sin(PI * 2 / len));
for (int i = 1; i < len; i++) g[i] = g[i - 1] * G;
}
void DFT(com *x, int op) {
for (int i = 0; i < len; i++) if (i < rev[i]) swap(x[i], x[rev[i]]);
for (int k = 2; k <= len; k <<= 1)
for (int j = 0; j < len; j += k)
for (int i = 0; i < k / 2; i++) {
com X = x[j + i], Y = x[j + i + k / 2] * g[len / k * (~op ? i : (i ? k - i : i))];
x[j + i] = X + Y, x[j + i + k / 2] = X - Y;
}
if (op < 0) for (int i = 0; i < len; i++) x[i].a /= len;
}
void mul(double *a, int n, double *b, int m, double *c) {
init(n + m);
static com F[MAX_LEN], G[MAX_LEN], S[MAX_LEN];
for (int i = 0; i < len; i++) F[i] = com(i <= n ? a[i] : 0.0, 0.0);
for (int i = 0; i < len; i++) G[i] = com(i <= m ? b[i] : 0.0, 0.0);
DFT(F, 1), DFT(G, 1);
for (int i = 0; i < len; i++) S[i] = F[i] * G[i];
DFT(S, -1);
for (int i = 0; i <= n + m; i++) c[i] = S[i].a;
}
}
const int maxn = 3e5 + 10;
int n;
double f[maxn], g[maxn], e[maxn];
int main() {
#ifdef xunzhen
freopen("force.in", "r", stdin);
freopen("force.out", "w", stdout);
#endif
n = read();
for (int i = 0; i < n; i++) scanf("%lf", &f[i]);
for (int i = 0; i < (n << 1); i++) if (i != n) g[i] = (i < n ? -1.0 : 1.0) / squ(i - n);
FFT::mul(f, n, g, (n << 1) - 1, e);
for (int i = n; i < (n << 1); i++) printf("%.3lf\n", e[i]);
return 0;
}
标签:__,Luogu3338,frac,sum,ZJOI2014,inline,define,getchar 来源: https://www.cnblogs.com/xunzhen/p/10352695.html