hdu 1402 A * B Problem Plus FFT

/*

hdu 1402 A * B Problem Plus FFT



这是我的第二道FFT的题



第一题是完全照着别人的代码敲出来的，也不明白是什么意思



这个代码是在前一题的基础上改的



做完这个题，我才有点儿感觉，原来FFT在这里就是加速大整数乘法而已



像前一题，也是一个大整数乘法，然后去掉一些非法的情况



*/

#pragma warning(disable : 4786)

#pragma comment(linker, "/STACK:102400000,102400000")

#include <iostream>

#include <string>

#include <cstring>

#include <cstdlib>

#include <cstdio>

#include <cmath>

#include <algorithm>

#include <functional>

#include <map>

#include <set>

#include <vector>

#include <stack>

#include <queue>//priority_queue

#include <bitset>

#include <complex>

#include <utility>

/*

**make_pair()

**/

using namespace std;

const double eps=1e-8;

const double PI=acos(-1.0);

const int inf=0x7fffffff;

template<typename T> inline T MIN(T a,T b){return a<b?a:b;}

template<typename T> inline T MAX(T a,T b){return a>b?a:b;}

template<typename T> inline T SQR(T a){return a*a;}

inline int dbcmp(double a){return a>eps?(1):(a<(-eps)?(-1):0);}



typedef __int64 LL;

int n;

const int size=400040;

int data[size/4];

int sum[size];

complex<double> x1[size],num1[size],num2[size];







void change(complex<double> 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<double> y[],int len,int on)

{

    change(y,len);

    for(int h = 2;h <= len;h <<= 1)

    {

        complex<double> wn(cos(-on*2*PI/h),sin(-on*2*PI/h));

        for(int j = 0;j < len;j += h)

        {

            complex<double> w(1,0);

            for(int k = j;k < j+h/2;k++)

            {

                complex<double> u = y[k];

                complex<double> 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].real(y[i].real()/len);

}

char s1[size];

char s2[size];

int main() {

	// your code goes here

	int t,i,j;

	int len,len1,len2;

	//cin>>t;

	while(gets(s1))

	{

		gets(s2);

		memset(num1,0,sizeof(num1));

		memset(num2,0,sizeof(num2));

		len1=strlen(s1);

		len2=strlen(s2);

		//printf("%d %d",len1,len2);

		len=1;

		while(len<2*len1||len<len2*2) len<<=1;

		for(i=len1-1,j=0;i>=0;--i,++j)

		{

			num1[j]=complex<double>(s1[i]-'0',0);

		}

		while(j<=len)

		{

			num1[j]=complex<double>(0,0);

			++j;

		}

		for(i=len2-1,j=0;i>=0;--i,++j)

		{

			num2[j]=complex<double>(s2[i]-'0',0);

		}

		while(j<=len)

		{

			num2[j]=complex<double>(0,0);

			++j;

		}

		fft(num1,len,1);

		fft(num2,len,1);

		for(i=0;i<len;++i)

		{

			num1[i]=num1[i]*num2[i];

		}

		fft(num1,len,-1);



		for(i=0;i<len;++i)

		{

			sum[i]=(int)(num1[i].real()+0.5);

		}

		for(i=0;i<len;++i)

		{

			sum[i+1]+=sum[i]/10;

			sum[i]=sum[i]%10;

		}

		len=len1+len2-1;

		while(sum[len]<=0&&len>0) --len;

		for(;len>=0;--len)

		{

			printf("%c",sum[len]+'0');

		}

		printf("\n");

	}

	return 0;

}