【bzoj2179】FFT快速傅立叶 FFT

卷积求法

1、DFT

2、点值相乘

3、逆DFT

DFT快速求法——FFT

1、按照下标分成奇数和偶数两个子序列

2、求出两个子序列的DFT

3、合并成本序列的DFT

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#define pi acos(-1)
#define maxn 200100

using namespace std;

struct abcd
{
double r,i;
abcd operator+(abcd x) {abcd ans=x;ans.r+=r;ans.i+=i;return ans;}
abcd operator-(abcd x) {abcd ans;ans.r=r-x.r;ans.i=i-x.i;return ans;}
abcd operator*(abcd x) {abcd ans;ans.r=r*x.r-i*x.i;ans.i=r*x.i+i*x.r;return ans;}
}a[maxn],b[maxn],p[maxn],temp[maxn];

int n,m,dight;
char s[maxn];
int st[maxn];

void FFT(abcd x[],int n,int type)
{
if (n==1) return;
for (int i=0;i<n;i+=2) temp[i>>1]=x[i],temp[i+n>>1]=x[i+1];
memcpy(x,temp,sizeof(abcd)*n);
abcd *l=x,*r=x+(n>>1);
FFT(l,n>>1,type);FFT(r,n>>1,type);
abcd root,w;
root.r=cos(2*pi*type/n);root.i=sin(2*pi*type/n);
w.r=1;w.i=0;
for (int i=0;i<n>>1;i++,w=w*root)
temp[i]=l[i]+w*r[i],temp[(n>>1)+i]=l[i]-w*r[i];
memcpy(x,temp,sizeof(abcd)*n);
}

int main()
{
scanf("%d",&n);
scanf("%s",s+1);
for (int i=0;i<n;i++) a[i].r=s[n-i]-'0';
scanf("%s",s+1);
for (int i=0;i<n;i++) b[i].r=s[n-i]-'0';
for (dight=1;dight<n<<1;dight<<=1);
FFT(a,dight,1);FFT(b,dight,1);
for (int i=0;i<dight;i++) p[i]=a[i]*b[i];
FFT(p,dight,-1);
int top=0;
for (int i=0;i<=n-1<<1;i++)
{
st[++top]+=(int)(p[i].r/dight+0.5);
st[top+1]+=st[top]/10;
st[top]%=10;
}
if (st[top+1]) top++;
for (int i=top;i>=1;i--) printf("%d",st[i]);printf("\n");
return 0;
}