bzoj 2179 FFT快速傅立叶

/**************************************************************
Problem: 2179
User: idy002
Language: C++
Result: Accepted
Time:1300 ms
Memory:10104 kb
****************************************************************/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long dnt;

const double eps = 1e-10;
const double dpi = 2*acos(-1);
const int N = (1<<17)+10;
const int P = 17;

struct Plex {
double x, y;
Plex(){}
Plex( double x, double y ):x(x),y(y){}
};
Plex operator-( const Plex &a ) {
return Plex(-a.x,-a.y);
}
Plex operator+( const Plex &a, const Plex &b ) {
return Plex(a.x+b.x,a.y+b.y);
}
Plex operator-( const Plex &a, const Plex &b ) {
return Plex(a.x-b.x,a.y-b.y);
}
Plex operator*( const Plex &a, const Plex &b ) {
return Plex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}

int n, p;
Plex a
, b
, c
, w[P+1];
Plex d
;

void init() {
for( int p=0,i=1; p<=P; p++,i<<=1 ) {
w[p] = Plex(cos(dpi/i),sin(dpi/i));
}
}
void read( Plex a
) {
static char buf
;
scanf( "%s", buf );
int len = strlen(buf);
reverse( buf, buf+len );
for( int i=0; buf[i]; i++ )
a[i].x = buf[i]-'0';
}
void print( int n, Plex a
) {
static int stk
, top;
for( int i=0; i<n; i++ ) {
stk[i] += int(a[i].x+0.49);
stk[i+1] += stk[i]/10;
stk[i] %= 10;
}
top = n-1;
while( top>=1 && stk[top]==0 ) top--;
if( top==0 ) {
printf( "0\n" );
} else {
for( int i=top; i>=0; i-- )
printf( "%d", stk[i] );
printf( "\n" );
}
}
int reverse( int a ) {
int b = 0;
for( int i=0; i<p; i++ )
if( a&(1<<i) ) b |= 1<<(p-1-i);
return b;
}
void fft( Plex a
, bool r ) {
for( int i=0; i<n; i++ ) {
int j=reverse(i);
if( i<j ) swap(a[i],a[j]);
}
for( int p=1; p<=::p; p++ ) {
for( int i=0; i<n; i+=(1<<p) ) {
Plex wo, wk;
int l = 1<<(p-1);
if( !r ) wo = w[p];
else wo = Plex(w[p].x,-w[p].y);
wk = w[0];
for( int j=0; j<l; j++,wk=wk*wo ) {
Plex lf = a[i+j], rg = a[i+j+l];
a[i+j] = lf + wk*rg;
a[i+j+l] = lf - wk*rg;
}
}
}
if( r ) for( int i=0; i<n; i++ )
a[i].x /= n;
}
int main() {
scanf( "%d", &n );
read(a);
read(b);
for( p=0; (1<<p)<n; p++ );
p++;
n = 1<<p;
init();
fft(a,0);
fft(b,0);
for( int i=0; i<n; i++ )
c[i] = a[i]*b[i];
fft(c,1);
print(n,c);
}