FFT（快速博立叶变换）

证明请参考算法概论的快速博立叶。

FFT递归：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5005;
const double PI = acos ( -1.0 );
struct node
{double real, image;node ( double a = 0, double b = 0 ) : real ( a ), image ( b ) { }node operator + ( const node &b ){return node ( real+b.real, image+b.image );}node operator - ( const node &b ){return node ( real-b.real, image-b.image );}node operator * ( const node &b ){return node ( real*b.real-image*b.image,real*b.image+image*b.real );}
};
node a[maxn], b[maxn];
int re[maxn];
node * FFT ( node x[maxn], int n, int dft )
{node * ans = ( node * ) malloc ( sizeof ( node )*n );if ( n == 1 ){ans[0] = x[0];return ans;}node lf[maxn], rt[maxn];for ( int i = 0; i < n; i += 2 )lf[i/2] = x[i];for ( int i = 1; i < n; i += 2 )rt[i/2] = x[i];node * L = FFT ( lf, n/2, dft );node * R = FFT ( rt, n/2, dft );node nw = node ( cos ( dft*2*PI/n ), sin ( dft*2*PI/n ) );node w = node ( 1, 0 );for ( int i = 0; i < n/2; i ++ ){ans[i] = L[i]+w*R[i];ans[i+n/2] = L[i]-w*R[i];w = w*nw;}return ans;
}
void solve ( )
{int n;scanf ( "%d", &n );for ( int i = 0; i < n; i ++ ){scanf ( "%lf", &a[i].real );a[i].image = 0;}for ( int i = 0; i < n; i ++ ){scanf ( "%lf", &b[i].real );b[i].image = 0;}int p = 1;while ( p < 2*n )p <<= 1;for ( int i = n; i < p; i ++ )a[i].real = a[i].image = b[i].real = b[i].image = 0;node * pa = FFT ( a, p, 1 );node * pb = FFT ( b, p, 1 );for ( int i = 0; i < p; i ++ )a[i] = pa[i]*pb[i];pa = FFT ( a, p, -1 );for ( int i = 0; i < p; i ++ )re[i] = pa[i].real/p+0.5;for ( int i = 0; i < p; i ++ )printf ( "%d ", re[i] );printf ( "\n" );
}
int main ( )
{solve ( );return 0;
}

FFT迭代代码（HDU1402）：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 500005;
const double PI = acos ( -1.0 );
struct node
{double real, image;node ( double a = 0, double b = 0 ) : real ( a ), image ( b ) { }node operator + ( const node &b ){return node ( real+b.real, image+b.image );}node operator - ( const node &b ){return node ( real-b.real, image-b.image );}node operator * ( const node &b ){return node ( real*b.real-image*b.image,real*b.image+image*b.real );}
};
node a[maxn], b[maxn], W[maxn];
int re[maxn];
char x[maxn], y[maxn];
void change ( node a[], int n )
{int p = 1;while ( ( 1 << p ) < n )p ++;for ( int i = 0; i < n; i ++ ){int j = 0, k = i, t = p;while ( t -- ){j <<= 1;j |= k&1;k >>= 1;}if ( j > i )swap ( a[i], a[j] );}
}
void FFT ( node a[], int n, int dft )
{change ( a, n );for ( int i = 1; i < n; i *= 2 ){node nw ( cos ( dft*PI*2/2/i ), sin ( dft*PI*2/2/i ) );for ( int j = 0; j < n; j += 2*i ){node w ( 1, 0 );for ( int k = 0; k < i; k ++ ){node tmp = a[j+k];node res = w*a[j+k+i];a[j+k] = tmp+res;a[j+k+i] = tmp-res;w = w*nw;}}}if ( dft == -1 ){for ( int i = 0; i < n; i ++ )a[i].real /= n;}
}
void solve ( )
{int lenx, leny;while ( ~ scanf ( "%s%s", x, y ) ){lenx = strlen ( x );for ( int i = 0; i < lenx; i ++ ){a[i].real = x[lenx-1-i]-'0';a[i].image = 0;}leny = strlen ( y );for ( int i = 0; i < leny; i ++ ){b[i].real = y[leny-1-i]-'0';b[i].image = 0;}int p = 1;while ( p < 2*lenx || p < 2*leny )p <<= 1;for ( int i = lenx; i < p; i ++ )a[i].real = a[i].image = 0;for ( int i = leny; i < p; i ++ )b[i].real = b[i].image = 0;FFT ( a, p, 1 );FFT ( b, p, 1 );for ( int i = 0; i < p; i ++ )a[i] = a[i]*b[i];FFT ( a, p, -1 );for ( int i = 0; i < p; i ++ )re[i] = a[i].real+0.5;for ( int i = 0; i < p; i ++ ){re[i+1] += re[i]/10;re[i] %= 10;}int h = 0;for ( int i = p-1; i >= 0; i -- )if ( re[i] ){h = i;break ;}for ( int i = h; i >= 0; i -- )printf ( "%d", re[i] );printf ( "\n" );}
}
int main ( )
{solve ( );return 0;
}

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