HDU 1402

http://acm.hdu.edu.cn/showproblem.php?pid=1402

fft做O(nlog(n))大数乘法，kuangbin的模板

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <math.h>

using namespace std;

const double PI = acos(-1.0);

//复数结构体

struct Complex

{

    double x,y;//实部和虚部 x+yi

    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);

    }

};

/*

* 进行FFT和IFFT前的反转变换。

* 位置i和 （i二进制反转后位置）互换

* len必须去2的幂

*/

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]);

        //交换互为小标反转的元素，i<j保证交换一次

        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的

        k = len/2;

        while(j >= k)

        {

            j -= k;

            k /= 2;

        }

        if(j < k)j += k;

    }

}

/*

* 做FFT

* len必须为2^k形式，

* on==1时是DFT，on==-1时是IDFT

*/

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 = 200010;

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);

        //求DFT

        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("%d",sum[i]);

        printf("\n");

    }

    return 0;

}

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