3527: [Zjoi2014]力
给出n个数qi，给出Fj的定义如下： 
令Ei=Fi/qi，求Ei.
化简一下，
FJ=Σ((qj)/(i-j)2)-Σ((qj)/(i-j)2)
另ti=1/i2 ,那么ti-j=1/(i-j)2。。。
那么Σ((qj)/(i-j)2)就变成了一个卷积，后面的把数组翻转一下就好了。。。
1 #include<iostream> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<cstdio> 6 #include<algorithm> 7 #include<string> 8 #include<map> 9 #include<queue> 10 #include<vector> 11 #include<set> 12 #define inf 1000000000 13 #define maxn 500000+5 14 #define maxm 10000+5 15 #define eps 1e-10 16 #define ll long long 17 #define for0(i,n) for(int i=0;i<=(n);i++) 18 #define for1(i,n) for(int i=1;i<=(n);i++) 19 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 20 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 21 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) 22 using namespace std; 23 int read(){ 24 int x=0,f=1;char ch=getchar(); 25 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 26 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 27 return x*f; 28 } 29 struct cp{ 30 double x,y; 31 cp operator +(cp b){return (cp){x+b.x,y+b.y};} 32 cp operator -(cp b){return (cp){x-b.x,y-b.y};} 33 cp operator *(cp b){return (cp){x*b.x-y*b.y,x*b.y+y*b.x};} 34 }; 35 double d[maxn],ans[maxn]; 36 const double PI=acos(-1.0); 37 cp a[maxn],b[maxn],c[maxn],y[maxn]; 38 int n,nn,m,len,rev[maxn]; 39 void fft(cp *x,int n,int flag){ 40 for0(i,n-1)y[rev[i]]=x[i]; 41 for0(i,n-1)x[i]=y[i]; 42 for(int m=2;m<=n;m<<=1){ 43 cp wn=(cp){cos(2.0*PI/m*flag),sin(2.0*PI/m*flag)}; 44 for(int i=0;i<n;i+=m){ 45 cp w=(cp){1.0,0};int mid=m>>1; 46 for0(j,mid-1){ 47 cp u=x[i+j],v=x[i+j+mid]*w; 48 x[i+j]=u+v;x[i+j+mid]=u-v; 49 w=w*wn; 50 } 51 } 52 } 53 if(flag==-1)for0(i,n-1)x[i].x/=n; 54 } 55 int main(){ 56 //freopen("input.txt","r",stdin); 57 //freopen("output.txt","w",stdout); 58 nn=n=read(); 59 n=2*n-1;m=1; 60 while(m<=n)m<<=1,len++;n=m; 61 for0(i,nn-1)scanf("%lf",&d[i]); 62 for0(i,n-1)a[i]=(cp){d[i],0}; 63 for0(i,nn-1)b[i]=i?(cp){1.0/((double)i*(double)i),0}:(cp){0,0}; 64 for2(i,nn,n-1)b[i]=(cp){0,0}; 65 for0(i,n-1){ 66 int x=i,y=0; 67 for1(j,len)y<<=1,y|=x&1,x>>=1; 68 rev[i]=y; 69 } 70 fft(a,n,1);fft(b,n,1); 71 for0(i,n-1)c[i]=a[i]*b[i]; 72 fft(c,n,-1); 73 for0(i,nn-1)ans[i]=c[i].x; 74 for0(i,nn-1)a[i]=(cp){d[nn-1-i],0}; 75 for2(i,nn,n-1)a[i]=(cp){0,0}; 76 for0(i,nn-1)b[i]=i?(cp){1.0/((double)i*(double)i),0}:(cp){0,0}; 77 for2(i,nn,n-1)b[i]=(cp){0,0}; 78 fft(a,n,1);fft(b,n,1); 79 for0(i,n-1)c[i]=a[i]*b[i]; 80 fft(c,n,-1); 81 for0(i,nn-1)ans[i]-=c[nn-1-i].x; 82 for0(i,nn-1)printf("%.5f\n",ans[i]); 83 return 0; 84 85 }
