洛谷 [P3723] 礼物

FFT

https://www.luogu.org/problemnew/solution/P3723
重点在于构造卷积的形式

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAXN = 400005;
const double PI = acos(-1);
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
struct Complex {
	double x, y;
	Complex(double xx = 0.0, double yy = 0.0) {
		x = xx; y = yy;
	}
	Complex operator + (const Complex &u) const {
		return Complex(x + u.x, y + u.y);
	}
	Complex operator - (const Complex &u) const {
		return Complex(x - u.x, y - u.y);
	}
	Complex operator * (const Complex &u) const {
		return Complex(x * u.x - y * u.y, x * u.y + y * u.x);
	}
}a[MAXN], b[MAXN];
int n, m, lim = 1, limcnt, rev[MAXN], num1[MAXN], num2[MAXN], ttt;
long long ans = 0, c;
void fft(Complex a[], int opt) {
	for(int i = 0; i <= lim; i++) {
		if(i < rev[i]) swap(a[i], a[rev[i]]);
	}
	for(int mid = 1; mid < lim; mid <<= 1) {
		Complex wn = Complex(cos(PI / mid), opt * sin(PI / mid));
		for(int R = mid << 1, j = 0; j < lim; j += R) {
			Complex w = Complex(1.0, 0.0);
			for(int k = 0; k < mid; k++) {
				Complex x = a[j + k], y = w * a[j + mid + k];
				a[j + k] = x + y;
				a[j + mid + k] = x - y;
				w = w * wn;
			}
		}
	}
	if(opt == -1) {
		for(int i = 0; i <= lim; i++) {
			a[i].x /= lim;
		}
	}
}
int main() {
	n = init(); m = init();
	for(int i = 1; i <= n; i++) {
		num1[i] = init();
		a[n - i].x = num1[i];
		ans += num1[i] * num1[i];
		ttt += num1[i];
	}
	for(int i = 0; i < n; i++) {
		num2[i] = init();
		b[i].x = b[i + n].x = num2[i];
		ans += num2[i] * num2[i];
		ttt -= num2[i];
	}
	double t = -(double)ttt / n;
	if(t > 0.0) c = (int)(t + 0.5);
	else c = (int) (t - 0.5);
	ans += n * c * c + 2 * c * ttt; 
	while(lim <= n * 3) lim <<= 1, limcnt++;
	for(int i = 0; i <= lim; i++)
		rev[i] = (rev[i>>1]>>1) | ((i&1) << (limcnt - 1)); 
	fft(a, 1); fft(b, 1);
	for(int i = 0; i <= lim; i++) {
		a[i] = a[i] * b[i];
	}
	fft(a, -1);
	int tmp = 0;
	for(int i = n - 1; i <= 2 * n - 1; i++) {
		tmp = max(tmp, (int)(a[i].x + 0.01));
	}
	ans -= 2 * tmp;
	cout << ans << endl;
	return 0;
}
posted @ 2018-05-24 09:56  Mr_Wolfram  阅读(53)  评论(0编辑  收藏