A * B Problem Plus HDU - 1402 （FFT）

Calculate A * B. 

1

2

1000

2

2

2000

题意：求A*B，A和B的长度都小于50000
题解：FFT的板子题，但FFT还不会，之后再贴一些想法

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<sstream>

#include<cmath>

#include<stack>

#include<map>

#include<cstdlib>

#include<vector>

#include<string>

#include<queue>

using namespace std;

#define ll long long

#define llu unsigned long long

#define INF 0x3f3f3f3f

const double PI = acos(-1.0);

const int maxn =  2e5+;

const int mod = 1e9+;

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=,j=len/;i<len-;i++)

    {

        if(i<j)

            swap(y[i],y[j]);

        k=len/;

        while(j>=k)

        {

            j -= k;

            k /= ;

        }

        if(j < k)

            j += k;

    }

}

void fft(Complex y[],int len,int on){

    change(y,len);

    for(int h=;h<=len;h <<= ){

        Complex wn(cos(-on *  * PI /h),sin(-on**PI/h));

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

            Complex w(,);

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

                Complex u = y[k];

                Complex t = w*y[k+h/];

                y[k] = u+t;

                y[k+h/] = u-t;

                w = w*wn;

            }

        }

    }

    if(on == -)

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

            y[i].x/=len;

}

Complex x1[maxn],x2[maxn];

char str1[maxn/],str2[maxn/];

int sum[maxn];

int main()

{

    while(scanf("%s",str1) != EOF)

    {

        scanf("%s", str2);

        int len1 = strlen(str1);

        int len2 = strlen(str2);

        int len = ;

        while(len < len1* || len < len2*)

            len<<=;

        for(int i=;i<len1;i++)

            x1[i] = Complex(str1[len1--i]-'',);

        for(int i=len1;i<len;i++)

            x1[i] = Complex(,);

        for(int i=;i<len2;i++)

            x2[i] = Complex(str2[len2--i]-'',);

        for(int i=len2;i<len;i++)

            x2[i] = Complex(,);

        fft(x1,len,);

        fft(x2,len,);

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

            x1[i] = x1[i]*x2[i];

        fft(x1,len,-);

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

            sum[i] = (int)(x1[i].x + 0.5);

        for(int i=;i<len;i++){

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

            sum[i] %= ;

        }

        len = len1+len2-;

        while(sum[len] <=  && len > )

            len--;

        for(int i=len;i>=;i--)

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

        printf("\n");

    }

}