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
*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;
}