P3803 FFT求多项式系数

P3803 FFT求多项式系数

传送门：https://www.luogu.org/problemnew/show/P3803

题意：

这是一道FFT模板题，求多项式系数

题解：

对a和b的系数求一个fft，转换为点乘式后

O（n）扫一遍直接算系数即可

对于多项式相加

\(\begin{array}{l}{A(x)=\left(x_{0}, y_{0}\right),\left(x_{1}, y_{1}\right) \ldots\left(x_{n}, y_{n}\right)} \\ {B(x)=\left(x_{0}, y_{0}^{\prime}\right),\left(x_{1}, y_{1}^{\prime}\right) \ldots .\left(x_{n}, y_{n}\right)}\end{array}\)

\(A(x)+B(x)=\left(x_{0}, y_{0}+y_{0}^{\prime}\right),\left(x_{1}, y_{1}+y_{1}^{\prime}\right)\left(x_{n}, y_{n}+y_{n}^{\prime}\right)\)

对于多项式相乘，我们需要补上一些项使得最后乘的的系数个数为2n+1

\(\begin{array}{l}{A(x)=\left(x_{0}, y_{0}\right),\left(x_{1}, y_{1}\right) \ldots\left(x_{2 n}, y_{2 n}\right)} \\ {B(x)=\left(x_{0}, y_{0}^{\prime}\right),\left(x_{1}, y_{1}^{\prime}\right) \ldots .\left(x_{2 n}, y_{2 n}\right)}\end{array}\)

\(A(x) B(x)=\left(x_{0}, y_{0} y_{0}^{\prime}\right),\left(x_{1}, y_{1} y_{1}^{\prime}\right)\left(x_{2 n}, y_{2 n} y_{2 n}^{\prime}\right)\)

代码：

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 1e7 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1.0);
LL quick_pow(LL x, LL y) {
    LL ans = 1;
    while(y) {
        if(y & 1) {
            ans = ans * x % mod;
        } x = x * x % mod;
        y >>= 1;
    } return ans;
}
struct complex {
    double x, y;
    complex(double xx = 0, double yy = 0) {
        x = xx, y = yy;
    }
} a[maxn], b[maxn];
complex operator + (complex a, complex b) {
    return complex(a.x + b.x, a.y + b.y);
}
complex operator - (complex a, complex b) {
    return complex(a.x - b.x, a.y - b.y);
}
complex operator * (complex a, complex b) {
    return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}

int n, m;
int l, r[maxn];
int limit = 1;
void fft(complex *A, int type) {
    for(int i = 0; i < limit; i++) {
        if(i < r[i]) swap(A[i], A[r[i]]);
    }
    for(int mid = 1; mid < limit; mid <<= 1) {
        complex Wn(cos(Pi / mid), type * sin(Pi / mid));
        for(int R = mid << 1, j = 0; j < limit; j += R) {
            complex w(1, 0);
            for(int k = 0; k < mid; k++, w = w * Wn) {
                complex x = A[j + k], y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + mid + k] = x - y;
            }
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    scanf("%d%d", &n, &m);
    for(int i = 0; i <= n; i++) scanf("%lf", &a[i].x);
    for(int i = 0; i <= m; i++) scanf("%lf", &b[i].x);
    while(limit <= n + m) limit <<= 1, l++;
    for(int i = 0; i < limit; i++) {
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    fft(a, 1);
    fft(b, 1);
    for(int i = 0; i <= limit; i++) {
        a[i] = a[i] * b[i];
    }
    fft(a, -1);
    for(int i = 0; i <= n + m; i++) {
        printf("%d ", (int)(a[i].x / limit + 0.5));
    }
    printf("\n");
    return 0;
}