acwing2241 礼物( FFT)-优快云博客

本文链接：https://blog.youkuaiyun.com/yusen_123/article/details/135831232

该篇文章介绍了一种使用快速傅立叶变换(FFT)算法处理复数数据的计算方法，用于求解两个复数序列的特定问题，并通过并行计算优化了计算过程，最终输出结果是两个复数序列的平方和减去相关系数的平方的最大整数值。

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题目：https://www.acwing.com/problem/content/2243/

思路：

代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
using namespace std;
#define LL long long
const int N = 4e5 + 10;
const LL mod = 998244353;
const double PI = acos(-1);
struct Complex
{
   double x, y;
   Complex operator +(const Complex& t) const
   {
       return { x + t.x,y + t.y };
   }
   Complex operator -(const Complex& t) const
   {
       return { x - t.x,y -t.y };
   }
   Complex operator *(const Complex& t) const
   {
       return { x*t.x - y*t.y,x*t.y+y*t.x};
   }
}a[N], b[N];
int n, m;
LL ans1, ans2;
int tot, bit;
int rel[N];
LL gg(int c,int bb)
{
   return (LL)n * c * c + 2 * c * bb;
}
void fft(Complex a[], int op)
{
   for (int i = 0; i < tot; i++)
       if (i < rel[i])swap(a[i], a[rel[i]]);
   for (int m = 2; m <= tot; m *= 2)
   {
       Complex w1 ={ cos(2 * PI / m),op * sin(2 * PI / m) };
       for (int i = 0; i < tot; i+=m)
       {
           Complex wk = { 1,0 };
           for (int j = 0; j < m / 2; j++)
           {
               Complex x = a[i + j], y = a[i + j + m / 2]*wk;
               a[i + j] = x + y;
               a[i + j + m / 2] = x - y;
               wk = w1 * wk;
           }
       }

}

}
int main() {
   cin >> n >> m;
   LL bb = 0;
   for (int i = 0; i <n; i++)
   {
       cin >> a[n - 1 - i].x;
       ans1 += (LL)a[n - 1 - i].x * a[n - 1 - i].x;
       bb +=(LL)a[n - 1 - i].x;
   }
   for (int i = 0; i < n; i++)
   {
       cin >> b[i].x;
       ans1 += b[i].x* b[i].x;
       bb -= b[i].x;
   }
   for (int i = 0; i <= n - 1; i++)
       a[i + n] = a[i];
   int cc = -bb / n;
   ans1 += min(gg(cc,bb), min(gg(cc+1,bb), gg(cc-1,bb)));
   n--;
   while ((1 << bit) < 3 * n + 1) bit++;
   tot = 1 << bit;
   for (int i = 0; i < tot; i++) rel[i] = rel[i / 2] / 2 + (i & 1 ? tot / 2 : 0);
   fft(a, 1);
   fft(b, 1);
   for (int i = 0; i < tot; i++) a[i] = a[i] * b[i];
   fft(a, -1);
   for (int i = n - 1; i <= 2 * (n - 1); i++)
   {
       ans2 = max((int)ans2, int(a[i].x / tot + 0.5));
   }
   ans1 = ans1 - 2 * ans2;
   cout << ans1 << endl;
   return 0;
}