# BZOJ 2179 [快速傅里叶变换 高精度乘法]

## 2179: FFT快速傅立叶

Time Limit: 10 Sec  Memory Limit: 259 MB
Submit: 3108  Solved: 1599
[Submit][Status][Discuss]

## Input

n<=60000

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=3e5+5;
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
const double PI=acos(-1);
struct Vector{
double x,y;
Vector(double a=0,double b=0):x(a),y(b){}
};
typedef Vector CD;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,Vector b){return Vector(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
Vector conj(Vector a){return Vector(a.x,-a.y);}

struct FastFourierTransform{
int n,rev[N];
CD omega[N],omegaInv[N];
void ini(int m){
n=1;
while(n<m) n<<=1;
for(int k=0;k<n;k++)
omega[k]=CD(cos(2*PI/n*k),sin(2*PI/n*k)),
omegaInv[k]=conj(omega[k]);

int k=0;
while((1<<k)<n) k++;
for(int i=0;i<n;i++){
int t=0;
for(int j=0;j<k;j++) if(i&(1<<j)) t|=(1<<(k-j-1));
rev[i]=t;
}
}
void transform(CD *a,CD *omega){
for(int i=0;i<n;i++) if(i<rev[i]) swap(a[i],a[rev[i]]);
for(int l=2;l<=n;l<<=1){
int m=l>>1;
for(CD *p=a;p!=a+n;p+=l)
for(int k=0;k<m;k++){
CD t=omega[n/l*k]*p[k+m];
p[k+m]=p[k]-t;
p[k]=p[k]+t;
}
}
}
void DFT(CD *a,int flag){
if(flag==1) transform(a,omega);
else{
transform(a,omegaInv);
for(int i=0;i<n;i++) a[i].x/=(double)n;
}
}
void FFT(CD *a,CD *b,int m){
ini(m);
DFT(a,1);DFT(b,1);
for(int i=0;i<n;i++) a[i]=a[i]*b[i];
DFT(a,-1);
}
}fft;

CD A[N],B[N];
int n,m,c[N];
char s1[N],s2[N];
int main(){
freopen("in","r",stdin);
scanf("%s%s",s1,s2);
for(int i=0;i<n;i++) A[i].x=s1[n-i-1]-'0',B[i].x=s2[n-i-1]-'0';
fft.FFT(A,B,m);
for(int i=0;i<m;i++) c[i]=int(A[i].x+0.5);//printf("c %d\n",c[i]);
for(int i=0;i<m;i++) c[i+1]+=c[i]/10,c[i]%=10;
while(c[m]) m++;
for(int i=m-1;i>=0;i--) printf("%d",c[i]);

return 0;
}
FFT 1880ms
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=3e5+5;
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
const double PI=acos(-1);
struct Vector{
double x,y;
Vector(double a=0,double b=0):x(a),y(b){}
};
typedef Vector CD;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,Vector b){return Vector(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}

struct FastFourierTransform{
int n,rev[N];
void ini(int m){
n=1;
while(n<m) n<<=1;

int k=0;
while((1<<k)<n) k++;
for(int i=0;i<n;i++){
int t=0;
for(int j=0;j<k;j++) if(i&(1<<j)) t|=(1<<(k-j-1));
rev[i]=t;
}
}
void DFT(CD *a,int flag){
for(int i=0;i<n;i++) if(i<rev[i]) swap(a[i],a[rev[i]]);
for(int l=2;l<=n;l<<=1){
int m=l>>1;
CD wn(cos(2*PI/l),flag*sin(2*PI/l));
for(CD *p=a;p!=a+n;p+=l){
CD w(1,0);
for(int k=0;k<m;k++){
CD t=w*p[k+m];
p[k+m]=p[k]-t;
p[k]=p[k]+t;
w=w*wn;
}
}
}
if(flag==-1) for(int i=0;i<n;i++) a[i].x/=n;
}
void FFT(CD *a,CD *b,int m){
ini(m);
DFT(a,1);DFT(b,1);
for(int i=0;i<n;i++) a[i]=a[i]*b[i];
DFT(a,-1);
}
}fft;
CD A[N],B[N];
int n,m,c[N];
char s1[N],s2[N];
int main(){
freopen("in","r",stdin);
scanf("%s%s",s1,s2);
for(int i=0;i<n;i++) A[i].x=s1[n-i-1]-'0',B[i].x=s2[n-i-1]-'0';
fft.FFT(A,B,m);
for(int i=0;i<m;i++) c[i]=int(A[i].x+0.5);//printf("c %d\n",c[i]);
for(int i=0;i<m;i++) c[i+1]+=c[i]/10,c[i]%=10;
while(c[m]) m++;
for(int i=m-1;i>=0;i--) printf("%d",c[i]);

return 0;
}

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=3e5+5;
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
ll P=1004535809,MOD=P;
ll Pow(ll a,ll b,ll MOD){
ll ans=1;
for(;b;b>>=1,a=a*a%MOD)
if(b&1) ans=ans*a%MOD;
return ans;
}
struct NumberTheoreticTransform{
int n,rev[N];
ll g;
void ini(int m){
n=1;
while(n<m) n<<=1;

int k=0;
while((1<<k)<n) k++;
for(int i=0;i<n;i++){
int t=0;
for(int j=0;j<k;j++) if(i&(1<<j)) t|=(1<<(k-j-1));
rev[i]=t;
}

g=3;
}
void DFT(ll *a,int flag){
for(int i=0;i<n;i++) if(i<rev[i]) swap(a[i],a[rev[i]]);
for(int l=2;l<=n;l<<=1){
int m=l>>1;
ll wn=Pow(g,flag==1?(P-1)/l:P-1-(P-1)/l,P);
for(ll *p=a;p!=a+n;p+=l){
ll w=1;
for(int k=0;k<m;k++){
ll t=w*p[k+m]%P;
p[k+m]=(p[k]-t+P)%P;
p[k]=(p[k]+t)%P;
w=w*wn%P;
}
}
}
if(flag==-1){
ll inv=Pow(n,P-2,P);;
for(int i=0;i<n;i++) a[i]=a[i]*inv%P;
}
}
void MUL(ll *A,ll *B){
DFT(A,1);DFT(B,1);
for(int i=0;i<n;i++) A[i]=A[i]*B[i]%MOD;
DFT(A,-1);
}
}fft;
int n,m,c[N];
char s1[N],s2[N];
ll A[N],B[N];
int main(){
freopen("in","r",stdin);
scanf("%s%s",s1,s2);
for(int i=0;i<n;i++) A[i]=s1[n-i-1]-'0',B[i]=s2[n-i-1]-'0';
fft.ini(m);
fft.MUL(A,B);
for(int i=0;i<m;i++) c[i]=A[i];//printf("c %d\n",c[i]);
for(int i=0;i<m;i++) c[i+1]+=c[i]/10,c[i]%=10;
while(c[m]) m++;
for(int i=m-1;i>=0;i--) printf("%d",c[i]);
}
NNT 3728ms