洛谷P3803 模板多项式乘法（FFT） fft

Posted 2020-10-21 Mychael

题目

这是一道FFT模板题

输入格式

给定一个n次多项式F(x)，和一个m次多项式G(x)。

请求出F(x)和G(x)的卷积。

输出格式

第一行2个正整数n,m。

接下来一行n+1个数字，从低到高表示F(x)的系数。

接下来一行m+1个数字，从低到高表示G(x))的系数。

输入样例

一行n+m+1个数字，从低到高表示F(x)?G(x)的系数。

输出样例

1 2
1 2
1 2 1

提示

1 4 5 2

题解

表示迭代还不是很懂
只好背模板。。。

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<complex>
#include<algorithm>
#define pi acos(-1)
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
using namespace std;
const int maxn = 3000005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57) {if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57) {out = (out << 3) + (out << 1) + c - ‘0‘; c = getchar();}
    return out * flag;
}
typedef complex<double> E;
E A[maxn],B[maxn];
int N,M,n,m,L,R[maxn];
void fft(E* a,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        E wn(cos(pi / i),f * sin(pi / i));
        for (int j = 0; j < n; j+= (i << 1)){
            E w(1,0);
            for (int k = 0; k < i; k++,w *= wn){
                E x = a[j + k],y = w * a[j + k + i];
                a[j + k] = x + y; a[j + k + i] = x - y;
            }
        }
    }
    if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
int main(){
    N = read(); M = read();
    for (int i = 0; i <= N; i++) A[i] = read();
    for (int i = 0; i <= M; i++) B[i] = read();
    m = N + M; for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    fft(A,1); fft(B,1);
    for (int i = 0; i <= n; i++) A[i] *= B[i];
    fft(A,-1);
    for (int i = 0; i <= m; i++) printf("%d ",(int)(A[i].real() + 0.1));
    return 0;
}