标签:max scan 直接 zjoi tchar opera double solution queue
给出n个数qi,
给出Fj的定义如下:\(F_j = \sum_{i<j}\frac{q_i q_j}{(i-j)^2 }-\sum_{i>j}\frac{q_i q_j}{(i-j)^2 }\)
令Ei=Fi/qi,求Ei.
\(n \le 10^5\)
\(F_j=\sum_{i=1}^{j-1} f(i)*g(j-i)-\sum_{i=j+1}^nf(i)*g(j-i)\)
\(F_j=\sum_{i=1}^{j-1} f(i)*g(j-i)-\sum_{i=1}^{n-j}f(n-i+1)*g(j-(n-i+1))\)
将\(f\)翻转
\(F_j=\sum_{i=1}^{j-1} f(i)*g(j-i)-\sum_{i=1}^{n-j}f(i)^{'}*g((n-j)-i+1)\)
直接FFT求卷积即可
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ls p<<1
#define rs p<<1|1
using namespace std;
typedef long long ll;
const int mxn=1e5+5;
const long double PI=acos(-1.0);
int n,l,lim=1,r[mxn];
inline int read() {
char c=getchar(); int x=0,f=1;
while(c>'9'||c<'0') {if(c=='-') f=-1;c=getchar();}
while(c<='9'&&c>='0') {x=(x<<3)+(x<<1)+(c&15);c=getchar();}
return x*f;
}
inline int chkmax(int &x,int y) {if(x<y) x=y;}
inline int chkmin(int &x,int y) {if(x>y) x=y;}
struct ed {
int to,nxt;
}t[mxn<<1];
struct cp {
long double x,y;
cp (long double xx=0,long double yy=0) {x=xx;y=yy;}
friend cp operator + (cp a,cp b) {
return cp(a.x+b.x,a.y+b.y);
}
friend cp operator - (cp a,cp b) {
return cp(a.x-b.x,a.y-b.y);
}
friend cp operator * (cp a,cp b) {
return cp(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}
}a[mxn],b[mxn],c[mxn];
void FFT(cp *p,int opt)
{
for(int i=0;i<lim;++i)
if(i<r[i]) swap(p[i],p[r[i]]);
for(int mid=1;mid<lim;mid<<=1) {
cp wn=cp(cos(PI/mid),opt*sin(PI/mid));
for(int len=mid<<1,j=0;j<lim;j+=len) {
cp w=cp(1,0);
for(int k=0;k<mid;++k,w=w*wn) {
cp x=p[j+k],y=w*p[j+mid+k];
p[j+k]=x+y; p[j+mid+k]=x-y;
}
}
}
}
int main()
{
n=read();
for(int i=1;i<=n;++i) {
scanf("%Lf",&a[i].x);
b[n-i+1].x=a[i].x;
c[i]=1.0/i/i;
}
while(lim<=n*2) lim<<=1,++l;
for(int i=0;i<lim;++i)
r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
FFT(a,1); FFT(b,1); FFT(c,1);
for(int i=0;i<=lim;++i) a[i]=a[i]*c[i];
for(int i=0;i<=lim;++i) b[i]=b[i]*c[i];
FFT(a,-1); FFT(b,-1);
for(int i=1;i<=n;++i)
printf("%.6Lf\n",a[i].x/lim+0.5-(b[n-i+1].x/lim+0.5));
return 0;
}
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标签:max scan 直接 zjoi tchar opera double solution queue
原文地址:https://www.cnblogs.com/list1/p/10504607.html