BZOJ2194: 快速傅立叶之二

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 200000+5
 26 
 27 #define maxm 20000000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 54 
 55     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 struct cp
 61 {
 62     double x,y;
 63     inline cp operator +(cp b){return (cp){x+b.x,y+b.y};}
 64     inline cp operator -(cp b){return (cp){x-b.x,y-b.y};}
 65     inline cp operator *(cp b){return (cp){x*b.x-y*b.y,x*b.y+y*b.x};}
 66 };
 67 const double PI=acos(-1.0);
 68 cp a[maxn],b[maxn],c[maxn],y[maxn];
 69 int n,m,len,rev[maxn],ans[maxn];
 70 char s[maxn];
 71 void fft(cp *x,int n,int flag)
 72 {
 73     for0(i,n-1)y[rev[i]]=x[i];
 74     for0(i,n-1)x[i]=y[i];//翻转
 75     for(int m=2;m<=n;m<<=1)
 76     {
 77         cp wn=(cp){cos(2.0*PI/m*flag),sin(2.0*PI/m*flag)};//w（m，1）
 78         for(int i=0;i<n;i+=m)//n个数按m个一份，每份内进行操作
 79         {
 80             cp w=(cp){1.0,0};int mid=m>>1;
 81             for0(j,mid-1)
 82             {
 83                 cp u=x[i+j],v=x[i+j+mid]*w;
 84                 x[i+j]=u+v;x[i+j+mid]=u-v;
 85                 w=w*wn;
 86             }//蝴蝶操作
 87         }
 88     }
 89     if(flag==-1)for0(i,n-1)x[i].x/=n;
 90 }
 91 
 92 int main()
 93 
 94 {
 95 
 96     freopen("input.txt","r",stdin);
 97 
 98     freopen("output.txt","w",stdout);
 99 
100     n=read();
101     for0(i,n-1)a[n-1-i].x=read(),b[i].x=read();
102     n=2*n-1;
103     m=1;
104     while(m<n)m<<=1,len++;swap(n,m);
105     for0(i,n-1)
106     {
107         int x=i,y=0;
108         for1(j,len)y<<=1,y|=(x&1),x>>=1;
109         rev[i]=y;
110     }
111     //for0(i,n-1)cout<<i<<‘ ‘<<a[i].x<<‘ ‘<<b[i].x<<endl;
112     fft(a,n,1);fft(b,n,1);
113     for0(i,n-1)c[i]=a[i]*b[i];
114     fft(c,n,-1);
115     //for0(i,n-1)cout<<i<<‘ ‘<<c[i].x<<endl;
116     for3(i,(m-1)>>1,0)printf("%lld\n",(ll)(c[i].x+0.5));
117 
118     return 0;
119 
120 }