codechef 3D Queries(MGCH3D)

资瓷点这里阅读QvQ

题外话

这个题是cc 9月challenge的最后一题，当时没仔细考虑，以为这是个近似计算的题。。。赛后看了题解才猛然醒悟

codechef 3D Queries(MGCH3D)

Description

求∑i!=j|A(Xi−Xj)+B(Yi−Yj)+C(Zi−Zj)+D|N(N−1)(Xi−Xj)4+(Yi−Yj)4+(Zi−Zj)4√\sum_{i!=j} \frac{|A(X_i - X_j) + B(Y_i - Y_j) + C(Z_i - Z_j) + D|}{N(N-1)\sqrt{(X_i-X_j)^4 + (Y_i-Y_j)^4+(Z_i-Z_j)^4}}

2≤N≤777777,1≤X,Y,Z,A,B,C≤772\le N \le 777777,1\le X, Y, Z, A, B, C\le 77, Q(1≤Q≤77)Q(1\le Q\le 77)组询问,每次给A,B,C,DA, B, C, D, 求上式的值。

Solution

x,y,zx, y, z坐标都很小(1≤x,y,z≤771\le x, y, z\le 77)，差的取值也很小，不妨构造多项式A[N2×x+N×y+z]+=1A[N^2\times x + N\times y + z] += 1, B[N2×(77−x)+N×(77−y)+(77−z)]+=1B[N^2\times (77-x)+N\times (77 - y)+(77-z)]+=1。令N=77∗2N=77*2

我们发现A×B=N2×(77+x[i]−x[j])+N×(77+y[i]−y[j])+(77+z[i]−z[j])A\times B=N^2\times (77+x[i]-x[j]) + N\times (77+y[i]-y[j]) + (77+z[i]-z[j])

显然x[i]−x[j],y[i]−y[j],z[i]−z[j]x[i]-x[j],y[i]-y[j],z[i]-z[j]我们可以通过，于是这题FFT一下就做完了= =

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> pii;
const int N = 153;
const double PI = acos(-1.0);
double D[80][80][80];
namespace FFT {
static const int W = 1 << 22;
static const double PI = acos(-1.0);
struct Complex {
double x, y;
Complex(double _x = 0, double _y = 0) : x(_x), y(_y) {}
Complex operator + (const Complex &t) const {return Complex(x + t.x, y + t.y);}
Complex& operator += (const Complex &t) {x += t.x, y += t.y;return *this;}
Complex operator - (const Complex &t) const {return Complex(x-t.x,y-t.y);}
Complex& operator -= (const Complex &t) {x -= t.x, y -= t.y;return *this;}
Complex operator * (const Complex &t) const {return Complex(x * t.x - y * t.y, x * t.y + y * t.x);}
Complex operator / (const double &t) const {return Complex(x / t,y / t);}
Complex& operator /= (const double &t) {x /= t, y /= t;return *this;}
double real() {return x;}
double imag() {return y;}
};
void fft(Complex a[], int n, int rev) {// rev=-1, reverse
for (int i = 1, j = 0, k; i < n; ++i) {
for (k = n >> 1; k > (j ^= k); k >>= 1);
if (i < j) swap(a[i], a[j]);
}
for (int s = 1, ss = 2; s < n; s <<= 1,ss <<= 1) {
Complex wn(cos(2 * PI * rev / ss), sin(2 * PI * rev / ss)), w;
for (int i = 0, j ; i < n; i += ss) {
for (j = i, w = 1; j < i + s; ++j, w = w * wn) {
Complex t = w * a[j + s];
a[j + s] = a[j] - t;
a[j] = a[j] + t;
}
}
}
if (rev == -1) for (int i = 0; i < n; ++i) a[i] /= n;
}
}
FFT::Complex A[FFT::W], B[FFT::W];
inline double sqrr(double x) {
return x * x * x * x;
}
int main() {
int n, q;
scanf("%d%d", &n, &q);
for (int i = 1, x, y, z; i <= n; ++i) {
scanf("%d%d%d", &x, &y, &z);
--x, --y, --z;
int t1 = N * N * x + N * y + z, t2 = N * N * (76 - x) + N * (76 - y) + (76 - z);
A[t1].x += 1, B[t2].x += 1;
}
for (int i = 0; i <= 77; ++i)
for (int j = 0; j <= 77; ++j)
for (int k = 0; k <= 77; ++k)
D[i][j][k] = sqrt(sqrr(i) + sqrr(j) + sqrr(k));
FFT::fft(A, FFT::W, 1);
FFT::fft(B, FFT::W, 1);
for (int i = 0; i < FFT::W; ++i)    A[i] = A[i] * B[i];
FFT::fft(A, FFT::W, -1);
while (q--) {
double ans = 0;
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = 0; i < FFT::W; ++i) {
int cnt = int(A[i].real() + 0.5);
if (!cnt)   continue;
int x = i / (N * N) - 76, y = i / N % N - 76, z = i % N - 76;
if (!x && !y && !z) continue;
ans += fabs(a * x + b * y + c * z + d) * cnt / D[abs(x)][abs(y)][abs(z)];
}
ans /= n, ans /= n - 1;
printf("%.10lf\n", ans);
}
return 0;
}

总结

自己还是太年轻，一看范围小就想着想忽略近似计算和乱搞，犯了方向性的错误= =。。还是要多加思考QvQ