## bzoj 2179 FFT快速傅立叶

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define db double
using namespace std;
const int N=6e4+5,M=N<<2; const db pi=acos(-1);
int n,len,r[M];
struct cpl{db x,y;int c;}a[M],b[M],I;
cpl operator+ (cpl a,cpl b){return (cpl){a.x+b.x,a.y+b.y};}
cpl operator- (cpl a,cpl b){return (cpl){a.x-b.x,a.y-b.y};}
cpl operator* (cpl a,cpl b){return (cpl){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
void fft(cpl *a,bool fx)
{
for(int i=0;i<len;i++)
if(i<r[i])swap(a[i],a[r[i]]);
for(int R=2;R<=len;R<<=1)//<=
{
int m=R>>1;
cpl Wn=(cpl){ cos(pi/m),fx?-sin(pi/m):sin(pi/m) };
for(int i=0;i<len;i+=R)
{
cpl w=I;
for(int j=0;j<m;j++,w=w*Wn)
{
cpl tmp=w*a[i+m+j];
a[i+m+j]=a[i+j]-tmp;
a[i+j]=a[i+j]+tmp;
}
}
}
}
int main()
{
scanf("%d",&n); n--; I.x=1;
for(int i=n,d;i>=0;i--)
{
scanf("%1d",&d); a[i].x=d;
}
for(int i=n,d;i>=0;i--)
{
scanf("%1d",&d); b[i].x=d;
}
len=1;
for(;len<=(n<<1);len<<=1);
for(int i=0;i<len;i++)
r[i]=(r[i>>1]>>1)+((i&1)?len>>1:0);
fft(a,0); fft(b,0);
for(int i=0;i<len;i++) a[i]=a[i]*b[i];
fft(a,1); int t=n<<1;
for(int i=0;i<=t;i++)
a[i].c=int(a[i].x/len+0.5);
for(int i=0;i<t;i++)//<
{
a[i+1].c+=a[i].c/10; a[i].c%=10;
}
while(!a[t].c&&t>=0)t--;
if(t<0)putchar('0');
for(int i=t;i>=0;i--)
printf("%d",a[i].c);
puts("");
return 0;
}

posted on 2018-11-26 15:12  Narh  阅读(80)  评论(0编辑  收藏  举报