hdu 1402 A * B Problem Plus[【FFT】

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
#include<queue>
#include<vector>
#include<list>
#include<bitset>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define N 200005
const double PI = acos(-1.0);
struct complex
{
double r,i;
complex(double _r = 0,double _i = 0)
{
r = _r; i = _i;
}
complex operator +(const complex &b)
{
return complex(r+b.r,i+b.i);
}
complex operator -(const complex &b)
{
return complex(r-b.r,i-b.i);
}
complex operator *(const complex &b)
{
return complex(r*b.r-i*b.i,r*b.i+i*b.r);
}
};
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]);
k = len/2;
while( j >= k)
{
j -= k;
k /= 2;
}
if(j < k)j += k;
}
}
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].r /= len;
}
complex x1
,x2
;
char a[N/4],b[N/4];
int num
;
int main()
{
while(~scanf("%s%s",a,b))
{
int len1=strlen(a);
int len2=strlen(b);
int l=1;
while(l<len1+len2) l<<=1;

for(int i=len1-1;i>=0;i--) x1[i]=complex(a[len1-i-1]-'0',0);
for(int i=len1;i<l;i++) x1[i]=complex(0,0);
for(int i=len2-1;i>=0;i--) x2[i]=complex(b[len2-i-1]-'0',0);
for(int i=len2;i<l;i++) x2[i]=complex(0,0);
fft(x1,l,1);
fft(x2,l,1);
for(int i=0;i<l;i++) x1[i]=x1[i]*x2[i];
fft(x1,l,-1);
for(int i=0;i<l;i++) num[i]=x1[i].r+0.5;
for(int i=0;i<l;i++)
{
num[i+1]+=num[i]/10;
num[i]%=10;
}
l=len1+len2-1;
int f=0;
for(int i=l;i>=0;i--)
{
if(!f&&num[i]==0) continue;
f=1;
printf("%d",num[i]);
}
if(!f) puts("0");
else puts("");
}
return 0;
}