hdu 1402 A * B Problem Plus(FFT)

FFT学习资料：http://blog.csdn.net/gyhguoge01234/article/details/76973859
初学fft，先找个不用思考，直接套模板的题来看看
我们可以把一个大数理解成a[0]+a[1]*10+a[2]*10^2+…+a[n]*10^n。把“10”当成未知数，这个多项式每一个次方项的系数就是大数每一数位上的数。然后dft，把系数表达式转换成点值表达式，表达式乘完后再idft，把点值表达式转换成系数表达式。

#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],str2[MAXN];int sum[MAXN];int main(){    while(scanf("%s %s",str1,str2) != EOF)    {        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("%d",sum[i]);        printf("\n");    }    return 0;}