BZOJ3527: [Zjoi2014]力

给出n个数qi，给出Fj的定义如下： 
 
令Ei=Fi/qi，求Ei.

题解：

为什么要乘qi再除以qi。。。

我们发现i-j+j=i为定值，也就是这是个卷积。可以分开算，前半部分

令 a[i]=1/(i*i),b[i]=q[i],然后它们的卷积c[i]就是前半部分的值。

后半部分我们直接把b翻转一下再来一次就可以了。

注意不要给a添加无用的和不存在的项。

代码：比原来的pascal快了10s 233

 

  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 500000+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     cp operator +(cp b){return (cp){x+b.x,y+b.y};}
 64     cp operator -(cp b){return (cp){x-b.x,y-b.y};}
 65     cp operator *(cp b){return (cp){x*b.x-y*b.y,x*b.y+y*b.x};}
 66 };
 67 double d[maxn],ans[maxn];
 68 //const double PI=3.1415926535898;
 69 const double PI=acos(-1.0);
 70 cp a[maxn],b[maxn],c[maxn],y[maxn];
 71 int n,nn,m,len,rev[maxn];
 72 void fft(cp *x,int n,int flag)
 73 {
 74     for0(i,n-1)y[rev[i]]=x[i];
 75     for0(i,n-1)x[i]=y[i];
 76     for(int m=2;m<=n;m<<=1)
 77     {
 78         cp wn=(cp){cos(2.0*PI/m*flag),sin(2.0*PI/m*flag)};
 79         for(int i=0;i<n;i+=m)
 80         {
 81             cp w=(cp){1,0};int mid=m>>1;
 82             for0(j,mid-1)
 83             {
 84                 cp u=x[i+j],v=x[i+j+mid]*w;
 85                 x[i+j]=u+v;x[i+j+mid]=u-v;
 86                 w=w*wn;
 87             }
 88         }
 89     }
 90     if(flag==-1)for0(i,n-1)x[i].x/=n;
 91 }
 92 
 93 int main()
 94 
 95 {
 96 
 97     freopen("input.txt","r",stdin);
 98 
 99     freopen("output.txt","w",stdout);
100 
101     nn=n=read();
102     n=2*n-1;m=1;
103     while(m<=n)m<<=1,len++;n=m;
104     for0(i,nn-1)scanf("%lf",&d[i]);
105     for0(i,n-1)a[i]=(cp){d[i],0};
106     for0(i,nn-1)b[i]=i?(cp){1.0/((double)i*(double)i),0}:(cp){0,0};
107     for2(i,nn,n-1)b[i]=(cp){0,0};
108     for0(i,n-1)
109     {
110         int x=i,y=0;
111         for1(j,len)y<<=1,y|=x&1,x>>=1;
112         rev[i]=y;
113     }
114     fft(a,n,1);fft(b,n,1);
115     for0(i,n-1)c[i]=a[i]*b[i];
116     fft(c,n,-1);
117     for0(i,nn-1)ans[i]=c[i].x;
118     for0(i,nn-1)a[i]=(cp){d[nn-1-i],0};
119     for2(i,nn,n-1)a[i]=(cp){0,0};
120     for0(i,nn-1)b[i]=i?(cp){1.0/((double)i*(double)i),0}:(cp){0,0};
121     for2(i,nn,n-1)b[i]=(cp){0,0};
122     fft(a,n,1);fft(b,n,1);
123     for0(i,n-1)c[i]=a[i]*b[i];
124     fft(c,n,-1);
125     for0(i,nn-1)ans[i]-=c[nn-1-i].x;
126     for0(i,nn-1)printf("%.5f\n",ans[i]);
127 
128     return 0;
129 
130 }
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posted @ 2014-12-16 16:07  ZYF-ZYF  Views(263)  Comments(0)  Edit  收藏  举报