bzoj 4827: [HNOI2017]礼物 （FFT)

Posted 2020-10-03 cminus

一道FFT

然而据说暴力可以水70分

然而我省选的时候看到了直接吓傻了  连暴力都没打

太弱了啊QAQ

emmmm

详细的拆开就看其他题解吧233

最后那一步卷积其实我一直没明白

后来画画图终于懂了

只要把其中一个反过来

多项式乘法的结果中的每一项系数就对应某一个Σx[i] * y[j] 的结果

前面几项是不完全的结果

但是太小了就被忽略啦

代码如下

/**************************************************************
    Problem: 4827
    User: cminus
    Language: C++
    Result: Accepted
    Time:5644 ms
    Memory:24568 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <complex>
using namespace std;
 
const int N = 500100;
typedef long long ll;
typedef complex<double> cp;
const double pi = acos(-1.0);
cp A[N], B[N];
 
void FFT(cp *y, int n, int type) {
    if (n == 1) return ;
    cp l[n >> 1], r[n >> 1];
    for (int i = 0; i <= n; i++)
        if (i & 1)  r[i >> 1] = y[i];
        else    l[i >> 1] = y[i];
    FFT(l, n >> 1, type); FFT(r, n >> 1, type);
    cp omegan(cos(2 * pi / n), sin(2 * pi * type / n)), omega(1, 0);
    for (int i = 0; i < n >> 1; i++) {
        y[i] = l[i] + r[i] * omega;
        y[i + (n >> 1)] = l[i] - r[i] * omega;
        omega *= omegan;
    }
}
 
int main() {
    int n, m, ans = 0, y = 0;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) {
        int x; scanf("%d", &x);
        A[n - i - 1] = x;
        ans += x * x;
    }
    for (int i = 0; i < n; i++) {
        int x; scanf("%d", &x);
        B[i] = x;
        ans += x * x;
        y += (int)B[i].real() - A[n - i - 1].real();
    }
    int n1;  for (n1 = 1; n1 <= n * 4; n1 <<= 1);
    for (int i = 0; i < n; i++)
        B[i + n] = B[i];
    FFT(A, n1, 1); FFT(B, n1, 1);
    for (int i = 0; i <= n1; i++)
        A[i] *= B[i];   
    FFT(A, n1, -1);
    int temp = 0, z = (-y) / n;
    for (int i = 0; i < n; i++)  temp = max(temp, (int)(A[i + n - 1].real() / n1 + 0.5));
    ans -= temp * 2;
    temp = z * z * n + y * z * 2;
    z += 1; temp = min(temp, z * z * n + y * z * 2);
    z -= 2; temp = min(temp, z * z * n + y * z * 2);
// 有理有据的精度优化
    ans += temp;
    printf("%d\n", ans);
    return 0;
}