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;
}