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("\n");

	}

	return 0;

}