BZOJ 2179 FFT快速傅立叶 ——FFT

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
 
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>
 
using namespace std;
 
#define maxn 200005
#define Complex cp
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i) 
#define mlog 16
#define inf (0x3f3f3f3f)
 
int read()
{
    int x=0,f=1; char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
 
struct cp{
    double x,y;//虚数可表示为 x+y*i  i^2=-1 
    cp operator + (cp a) {return (cp){x+a.x,y+a.y};}
    cp operator - (cp a) {return (cp){x-a.x,y-a.y};}
    cp operator * (cp a) {return (cp){x*a.x-y*a.y,x*a.y+y*a.x};}
}a[maxn],b[maxn],c[maxn];
 
double pi=acos(-1.0); // π 
int n,m,rev[maxn],len,ans[maxn];
char s[maxn];

void FFT(Complex * x,int n,int f)
{
	F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]); //构造迭代的形式
	for (int m=2;m<=n;m<<=1)
	{
		Complex wn=(Complex){cos(2.0*pi/m*f),sin(2.0*pi/m*f)}; //当前的主单位根
		for (int i=0;i<n;i+=m)
		{
			Complex w=(Complex){1.0,0}; 
			for (int j=0;j<(m>>1);++j)
			{
				Complex u=x[i+j],v=x[i+j+(m>>1)]*w;//计算对应的多项式的值 
				x[i+j]=u+v; x[i+j+(m>>1)]=u-v;
				w=w*wn;//在复数域中旋转一个角度 
			}
		}
	}
} 
 
int main()
{
    n=read();
    scanf("%s",s); F(i,0,n-1) a[i].x=s[n-1-i]-‘0‘;
    scanf("%s",s); F(i,0,n-1) b[i].x=s[n-1-i]-‘0‘;
    m=1; n=2*n-1;
    while (m<=n) m<<=1,len++; n=m;
    F(i,0,n-1)
    {
        int t=i,ret=0;
        F(j,1,len) ret<<=1,ret|=t&1,t>>=1;
        rev[i]=ret;
    }//二进制翻转，便于化分治为循环迭代 
    FFT(a,n,1); FFT(b,n,1); //FFT 
    F(i,0,n-1) c[i]=a[i]*b[i];
    FFT(c,n,-1); //IFFT
    F(i,0,n-1) ans[i]=(c[i].x/n)+0.5;//精度QAQ 
    F(i,0,n-1) ans[i+1]+=ans[i]/10,ans[i]%=10;//进位QwQ 
    n++;
    while (!ans[n]&&n) n--;
    D(i,n,0) printf("%d",ans[i]);
}