BZOJ2194: 快速傅立叶之二

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2194

题目大意：请计算C[k]=sigma(a[i]*b[i-k]) 其中 k < = i < n ，并且有 n < = 10 ^ 5。 a,b中的元素均为小于等于100的非负整数。

题解：这就是所谓的卷积，找个时间一定要好好看看，上FFT咯

代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cstdio>
 5 #include<cmath>
 6 #define maxn 270005
 7 #define pi acos(-1)
 8 using namespace std;
 9 int n,m;
10 int ans[maxn],rev[maxn];
11 int l;
12 struct F{
13     double rea,ima;
14     F operator +(const F &x){return (F){rea+x.rea,ima+x.ima};}
15     F operator -(const F &x){return (F){rea-x.rea,ima-x.ima};}
16     F operator *(const F &x){return (F){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}
17 }a[maxn],b[maxn],c[maxn],w,wn,t1,t2;
18 int read()
19 {
20     int x=0; char ch; bool bo=0;
21     while (ch=getchar(),ch<‘0‘||ch>‘9‘) if (ch==‘-‘) bo=1;
22     while (x=x*10+ch-‘0‘,ch=getchar(),ch>=‘0‘&&ch<=‘9‘);
23     if (bo) return -x; return x;
24 }
25 void fft(F *a,int type)
26 {
27     for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
28     for (int s=2; s<=n; s<<=1)
29     {
30         wn=(F){cos(type*2*pi/s),sin(type*2*pi/s)};
31         for (int i=0; i<n; i+=s)
32         {
33             w=(F){1,0};
34             for (int j=i; j<(i+(s>>1)); j++,w=w*wn)
35             {
36                 t1=a[j],t2=a[j+(s>>1)]*w;
37                 a[j]=t1+t2,a[j+(s>>1)]=t1-t2;
38             }
39         }
40     }
41 }
42 int re(int x)
43 {
44     int t=0;
45     for (int i=0; i<l; i++) t<<=1,t|=x&1,x>>=1;
46     return t;
47 }
48 void init()
49 {
50     m=read();
51     for (n=1; n<2*m; n<<=1) l++;
52     for (int i=0; i<m; i++)
53     {
54         a[i].rea=read(),b[m-i-1].rea=read();
55     }
56     for (int i=0; i<n; i++) rev[i]=re(i);
57     fft(a,1); fft(b,1);
58     for (int i=0; i<n; i++) c[i]=a[i]*b[i];
59     fft(c,-1);
60     for (int i=0; i<n; i++) ans[i]=(int)(c[i].rea/n+0.5);
61     for (int i=m-1; i<2*m-1; i++) printf("%d\n",ans[i]);
62
63 }
64 int main()
65 {
66     init();
67 }