[FFT] 51Nod 1028 大数乘法 V2

FFT版子题

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<complex>
#include<algorithm>
#include<cstring>
#define PI acos(-1.0)
using namespace std;
typedef complex<double> E;

const int N=500005;

int n,m,L;
char s
;
E a
,b
;
int c
,R
;

inline void FFT(E *a,int r){
for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);
for (int i=1;i<n;i<<=1){
E wn(cos(PI/i),r*sin(PI/i));
for (int j=0;j<n;j+=(i<<1)){
E w(1,0);
for (int k=0;k<i;k++,w*=wn){
E x=a[j+k],y=w*a[j+k+i];
a[j+k]=x+y; a[j+k+i]=x-y;
}
}
}
if (r==-1) for (int i=0;i<n;i++) a[i]/=n;
}

int main(){
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
int len1,len2;
scanf("%s",s); len1=strlen(s); while (getchar()!='\n');
for (int i=1;i<=len1;i++) a[i-1]=s[len1-i]-'0';
scanf("%s",s); len2=strlen(s);
for (int i=1;i<=len2;i++) b[i-1]=s[len2-i]-'0';
m=len1+len2;
for (n=1;n<=m;n<<=1) L++;
for(int i=0;i<n;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
FFT(a,1); FFT(b,1);
for (int i=0;i<n;i++) a[i]*=b[i];
FFT(a,-1);
for(int i=0;i<=m;i++) c[i]=(int)(a[i].real()+0.1);
for(int i=0;i<=m;i++)
if(c[i]>=10){
c[i+1]+=c[i]/10,c[i]%=10;
if(i==m) m++;
}
while (!c[m]) m--;
for (int i=m;i>=0;i--) printf("%d",c[i]);
return 0;
}