BZOJ2194: 快速傅立叶之二

给A,B,求C,$C_k=\sum_{i=0}^{n-1}A_iB_{i-k}$。

B反过来就是模板了。能不能算一道题呢?

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<algorithm>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<complex>
 7 #include<math.h>
 8 //#include<iostream>
 9 using namespace std;
10 
11 int n,m,wei;
12 #define maxn 262222
13 const double pi=acos(-1);
14 typedef complex<double> cp;
15 cp a[maxn],b[maxn],c[maxn]; int rev[maxn];
16 
17 void dft(cp *a,int n,int type)
18 {
19     for (int i=0;i<n;i++) if (i<rev[i]) {cp t=a[i]; a[i]=a[rev[i]]; a[rev[i]]=t;}
20     for (int i=1;i<n;i<<=1)
21     {
22         cp base=cp(cos(pi/i),type*sin(pi/i));
23         for (int j=0,p=i<<1;j<n;j+=p)
24         {
25             cp t=cp(1,0);
26             for (int k=0;k<i;k++,t=t*base)
27             {
28                 cp tmp=t*a[j+k+i];
29                 a[j+k+i]=a[j+k]-tmp;
30                 a[j+k]+=tmp;
31             }
32         }
33     }
34 }
35 
36 void mul(cp *a,cp *b,cp *c)
37 {
38     for (int i=0;i<n;i++)
39     {
40         rev[i]=0;
41         for (int j=0;j<wei;j++) rev[i]|=((i>>j)&1)<<(wei-j-1);
42     }
43     dft(a,n,1); dft(b,n,1); for (int i=0;i<n;i++) c[i]=a[i]*b[i];
44     dft(c,n,-1); for (int i=0;i<n;i++) c[i]/=n;
45 }
46 
47 int main()
48 {
49     scanf("%d",&n); n--;
50     for (int i=0,x,y;i<=n;i++) scanf("%d",&x),a[i]=x,scanf("%d",&y),b[n-i]=y;
51     m=n+n; for (n=1,wei=0;n<=m;n<<=1,wei++); mul(a,b,c);
52     for (int i=(m>>1);i<=m;i++) printf("%d\n",(int)(c[i].real()+0.5));
53     return 0;
54 }
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posted @ 2018-02-06 12:31  Blue233333  阅读(88)  评论(0编辑  收藏  举报