HihoCoder1388(Periodic Signal)-fft(快速傅里叶变换)

最新推荐文章于 2024-01-15 15:43:59 发布

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本文介绍了一种使用快速傅立叶变换（FFT）解决周期信号问题的方法。通过构造两个序列并计算它们的卷积，可以有效地找到问题的解。这种方法的时间复杂度为O(n*log(n))，能够高效地处理大规模数据。

题目链接：Periodic Signal

题意：求

思路：

求题目所给的最小值转化为了求的最大值。

构造序列两个

求这两个序列的卷积，就可以得到，k = 0, 1, 2, ..., n - 1，

求卷积用fft来实现，复杂度为o(n*log(n))

由于用fft算出来的数精度误差会比较大，可以用这个方法先算出k的位置，然后将k代入原始式子，求得答案。

代码：

# pragma comment(linker, "/STACK:1024000000,1024000000")
# include <iostream>
# include <algorithm>
# include <cstdio>
# include <cstring>
# include <cmath>
using namespace std;
typedef long long ll;
const long double PI = acos(-1.0);
const int maxn = 5e5 + 5;
int a[maxn], b[maxn];
int n;

struct Complex {
    long double x, y;
    Complex() : x(0.0), y(0.0) { }
    Complex(long double x, long double y) : x(x), y(y) { }
    Complex operator - (const Complex &b) const {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator + (const Complex &b) const {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator * (const Complex &b) const {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
} A[maxn], B[maxn];

void change(Complex y[], int len) {
    int i, j, k;
    for (i = 1, j = len / 2; i < len - 1; i++) {
        if (i < j) swap(y[i], y[j]);
        k = len / 2;
        while (j >= k) {
            j -= k; k /= 2;
        }
        if (j < k) j += k;
    }
}

void fft(Complex y[], int len, int on) {
    change(y, len);
    for (int h = 2; h <= len; h <<= 1) {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for (int j = 0; j < len; j += h) {
            Complex w(1, 0);
            for (int k = j; k < j + h / 2; k++) {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }
    if(on == -1)
    for (int i = 0; i < len; i++) y[i].x /= len;
}

ll sqr(ll x) {
    return x * x;
}

int main(void)
{
    int T; scanf("%d", &T);
    while (T-- && scanf("%d", &n)) {
        for (int i = 0; i < n; ++i) {
            scanf("%d", a + i);
            A[i] = Complex(a[i] * 1.0, 0.0);
        }
        for (int i = 0; i < n; ++i) {
            scanf("%d", b + i);
            B[n - i - 1] = Complex(b[i] * 1.0, 0.0);
        }
        for (int i = n; i < n * 2; ++i) B[i] = B[i - n], A[i] = Complex(0.0, 0.0);
        n *= 2;
        int len = 1;
        while (len < n) len <<= 1;
        len <<= 1;
        for (int i = n; i < len; ++i) A[i] = B[i] = Complex(0.0, 0.0);
        fft(A, len, 1); fft(B, len, 1);
        for (int i = 0; i < len; ++i) A[i] = A[i] * B[i];
        fft(A, len, -1);
        int k = 0;
        long double Max = A[n * 2 - 1].x;
        n /= 2;
        for (int i = n * 2 - 2; i >= n; --i) {
            if (Max < A[i].x) {
                Max = A[i].x; k = n * 2 - i - 1;
            }
        }
        ll ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += sqr(a[i] - b[(i + k) % n]);
        }
        printf("%lld\n", ans);
    }

    return 0;
}