HDU 1402 A * B Problem Plus FFT

跟着Bin哥一起学。

#include <algorithm>
#include <math.h>
#include <cstdio>
#include <iostream>
#include <string.h>

using namespace std;

const double PI=acos(-1.0);

struct Complex
{
	double x,y;
	Complex(double _x=0.0,double _y=0.0)
	{
		x=_x;
		y=_y;
	}
	Complex operator -(const Complex &b)const
	{
		return Complex(x-b.x,y-b.y);
	}
	Complex operator +(const Complex &b)const
	{
		return Complex(x+b.x,y+b.y);
	}
	Complex operator *(const Complex &b)const
	{
		return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
	}
};

void change(Complex y[],int len)
{
	int i,j,k;
	for(i=1,j=len/2;i<len-1;i++)
	{
		if(i<j)	swap(y[i],y[j]);
		k=len/2;
		while(j>=k)
		{
			j-=k;
			k/=2;
		}
		if(j<k)	j+=k;
	}
}

void fft(Complex y[],int len,int on)
{
	change(y,len);
	for(int h=2;h<=len;h<<=1)
	{
		Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
		for(int j=0;j<len;j+=h)
		{
			Complex w(1,0);
			for(int k=j;k<j+h/2;k++)
			{
				Complex u=y[k];
				Complex t=w*y[k+h/2];
				y[k]=u+t;
				y[k+h/2]=u-t;
				w=w*wn;
			}
		}
	}
	if(on==-1)
		for(int i=0;i<len;i++)
			y[i].x/=len;
}

const int maxn = 200000 + 10;
Complex x1[maxn],x2[maxn];
char str1[maxn/2],str2[maxn/2];
int sum[maxn];

int main()
{
	while(~scanf("%s%s",str1,str2))
	{
		int len1=strlen(str1);
		int len2=strlen(str2);
		int len=1;
		while(len<len1*2||len<len2*2)  len<<=1;
		for(int i=0;i<len1;i++)
			x1[i]=Complex(str1[len1-1-i]-'0',0);
		for(int i=len1;i<len;i++)
			x1[i]=Complex(0,0);
		for(int i=0;i<len2;i++)
			x2[i]=Complex(str2[len2-1-i]-'0',0);
		for(int i=len2;i<len;i++)
			x2[i]=Complex(0,0);
		fft(x1,len,1);
		fft(x2,len,1);
		for(int i=0;i<len;i++)
			x1[i]=x1[i]*x2[i];
		fft(x1,len,-1);
		for(int i=0;i<len;i++)
			sum[i]=(int)(x1[i].x+0.5);
		for(int i=0;i<len;i++)
		{
			sum[i+1]+=sum[i]/10;
			sum[i]%=10;
		}
		len=len1+len2-1;
		while(sum[len]<=0&&len>0)  len--;
		for(int i=len;i>=0;i--)
			printf("%c",sum[i]+'0');
		printf("
");
	}
	return 0;
}