BZOJ4827: [Hnoi2017]礼物(FFT 二次函数)

题意

题目链接

Sol

越来越菜了。。裸的FFT写了1h。。

思路比较简单，直接把

\(\sum (x_i - y_i + c)^2\)

拆开

发现能提出一坨东西，然后与c有关的部分是关于C的二次函数可以直接算最优取值

剩下的要求的就是\(max (\sum x_i y_i)\)

画画图就知道把y序列倒过来就是个裸的FFT了。

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

#define LL long long

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;

const double eps = 1e-9, Pi = acos(-1);

template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}

template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}

template <typename A> inline void debug(A a){cout << a << '\n';}

template <typename A> inline LL sqr(A x){return 1ll * x * x;}

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, M;

int a[MAXN], b[MAXN], c[MAXN], t[MAXN], nx, ny, rev[MAXN];

double sx, sy;

struct com {

	double x, y;

}A[MAXN], B[MAXN], C[MAXN];

com operator * (const com a, const com b) {

	return (com) {a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};

}

com operator + (const com a, const com b) {

	return (com) {a.x + b.x, a.y + b.y};

}

com operator - (const com a, const com b) {

	return (com) {a.x - b.x, a.y - b.y};

}

void FFT(com *A, int lim, 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) {

		com Wn = (com) {cos(Pi / mid), opt * sin(Pi / mid)};

		for(int i = 0, R = mid << 1; i <= lim; i += R) {//tag

			com w = (com) {1, 0};

			for(int j = 0; j < mid; j++, w = w *  Wn) {

				com x = A[i + j], y = w * A[i + j + mid];

				A[i + j] = x + y;

				A[i + j + mid] = x - y;

			}

		}

	}

	if(opt == -1) {

		for(int i = 0; i <= lim; i++) A[i].x /= lim;

	}

}

LL check(int c) {

	ny = 0;

	memcpy(b, t, sizeof(t));

	for(int i = 0; i < N; i++) b[i] += c, b[i + N] = b[i], ny += b[i] * b[i];

	int M = 2 * N - 1;

	reverse(b, b + M + 1);

	memset(A, 0, sizeof(A));

	memset(B, 0, sizeof(B));

	memset(C, 0, sizeof(C));

	N--;

	for(int i = 0; i <= N; i++) A[i].x = a[i];

	for(int i = 0; i <= M; i++) B[i].x = b[i];

	int lim = 1, len = 0;

	while(lim <= N + M) lim <<= 1, len++;

	for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (len - 1));

	FFT(A, lim, 1); FFT(B, lim, 1);

	for(int i = 0; i <= lim; i++) C[i] = A[i] * B[i];

	FFT(C, lim, -1);

	N++;

	LL ret = 0;

	for(int i = 0; i <= M; i++) chmax(ret, C[i].x + 0.5);

	return -2 * ret + nx + ny;

}

signed main() {

	N = read(); M = read();

	for(int i = 0; i < N; i++) a[i] = read(), sx += a[i], nx += a[i] * a[i];

	for(int i = 0; i < N; i++) b[i] = read(), sy += a[i];

	memcpy(t, b, sizeof(b));

	int c = - (sx - sy) / N;

	LL ans = check(c);

	ans = min(ans, min(check(c - 1), check(c + 1)));

	cout << ans;

    return 0;

}

巴特西

BZOJ4827: [Hnoi2017]礼物(FFT 二次函数)

题意

Sol

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