题目大意：给定两个高精度整数，求两个数的乘积

FFT大法好

系统的complex比手写慢了2.5倍 简直吓死人- -

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 131080
#define PI 3.1415926535897932384626433832795028841971
using namespace std;
namespace PoPoQQQ_Complex{
	template<typename T>class Complex{
	public:
		T real,imaginary;
		Complex() {}
		Complex(T _,T __=0.0):real(_),imaginary(__) {}
		Complex& operator += (const Complex<T> &x) { real+=x.real;imaginary+=x.imaginary;return *this; }
		Complex& operator -= (const Complex<T> &x) { real-=x.real;imaginary-=x.imaginary;return *this; }
		friend Complex operator + (const Complex<T> &x,const Complex<T> &y) { Complex re=x;re+=y;return re; }
		friend Complex operator - (const Complex<T> &x,const Complex<T> &y) { Complex re=x;re-=y;return re; }
		friend Complex operator * (const Complex<T> &x,const Complex<T> &y)
		{ return Complex(x.real*y.real-x.imaginary*y.imaginary,x.real*y.imaginary+x.imaginary*y.real); }
		Complex& operator *= (const Complex &x) { return *this=(*this)*x; }
	};
}
using namespace PoPoQQQ_Complex;
typedef Complex<double> abcd;
int n;
abcd a[M],b[M],p[M];
char s[M];
void FFT(abcd x[],int n,int type)
{
	static abcd temp[M];
	int i;
	if(n==1) return ;
	for(i=0;i<n;i+=2)
		temp[i>>1]=x[i],temp[i+n>>1]=x[i+1];
	memcpy(x,temp,sizeof(abcd)*n);
	abcd *l=x,*r=x+(n>>1);
	FFT(l,n>>1,type);FFT(r,n>>1,type);
	abcd root(cos(2*type*PI/n),sin(2*type*PI/n) ),w(1);
	for(i=0;i<n>>1;i++,w*=root)
		temp[i]=l[i]+w*r[i],temp[i+(n>>1)]=l[i]-w*r[i];
	memcpy(x,temp,sizeof(abcd)*n);
}
int main()
{
	int i,digit;
	cin>>n;
	scanf("%s",s+1);
	for(i=0;i<n;i++)
		a[i].real=s[n-i]-'0';
	scanf("%s",s+1);
	for(i=0;i<n;i++)
		b[i].real=s[n-i]-'0';
	for(digit=1;digit<n<<1;digit<<=1);
	FFT(a,digit,1);FFT(b,digit,1);
	for(i=0;i<digit;i++)
		p[i]=a[i]*b[i];
	FFT(p,digit,-1);
	static int stack[M],top;
	for(i=0;i<=n-1<<1;i++)
	{
		++top;
		stack[top]+=int(p[i].real/digit+0.5);
		stack[top+1]+=stack[top]/10;
		stack[top]%=10;
	}
	if(stack[top+1]) ++top;
	while(top)
		putchar(stack[top--]+'0');
	puts("");
	return 0;
}

BZOJ 2179 FFT快速傅立叶 快速傅里叶变换