[bzoj3509][CodeChef]COUNTARI

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题意：给你n个数ai，求有多少个数对(i,j,k)满足$1leqslant i<j<kleqslant n$且aj*2=ai+ak    n<=100000 ai<=30000

题解：考虑构造一个生成函数，只要把左右的生成函数乘起来，然后枚举i就行了。但是每次平方都需要$nlogn$的时间，总复杂度$n^{2}logn$，不能过。

考虑分块，块外的(即满足$1leqslant i<Lleqslant jleqslant R<kleqslant n$的)用生成函数处理一下，块内的直接暴力算.

块的大小是k的时候，复杂度$k^{2}*frac{n}{k}+frac{n}{k}nlogn$，得出k大概等于$sqrt{nlogn}$时候最小，复杂度$nsqrt{nlogn}$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define getchar() (*S++)
#define MN 100000
#define pi acos(-1)
#define ll long long
char B[1<<26],*S=B;
using namespace std;
int X;char ch;
inline int read()
{
    X = 0 , ch = getchar();
    while(ch < '0' || ch > '9') ch = getchar();
    while(ch >= '0' && ch <= '9'){X = X * 10 + ch - '0';ch = getchar();}
    return X;
}

struct cp{
    double r,u;
    cp(double _r=0,double _u=0):r(_r),u(_u){}
    cp operator+(cp b){return cp(r+b.r,u+b.u);}
    cp operator-(cp b){return cp(r-b.r,u-b.u);}
    cp operator*(cp b){return cp(r*b.r-u*b.u,r*b.u+u*b.r);}
    cp operator/(double y){return cp(r/y,u/y);}
}w[2][MN],a[MN],b[MN];

int n,s[MN+5],size,N;ll num1[MN],num2[MN];
ll ans=0;

void init(int mx)
{
    w[0][0]=w[1][mx]=cp(1,0);
    w[0][1]=w[1][mx-1]=cp(cos(2*pi/mx),sin(2*pi/mx));
    for(int i=2;i<=mx;i++)
        w[0][i]=w[1][mx-i]=w[0][i-1]*w[0][1];
}

void fft(cp*x,int b)
{
    for(register int i=0,j=0;i<N;++i)
    {
        if(i>j)swap(x[i],x[j]);
        for(int k=N>>1;(j^=k)<k;k>>=1);
    }
    for(register int i=2;i<=N;i<<=1)for(register int j=0;j<N;j+=i)for(register int k=0;k<i>>1;k++)
    {
        cp t=x[j+k+(i>>1)]*w[b][N/i*k];
        x[j+k+(i>>1)]=x[j+k]-t;
        x[j+k]=x[j+k]+t;
    }
    if(b)for(register int i=0;i<N;i++)x[i]=x[i]/N;
}

int main()
{
    fread(B,1,1<<26,stdin);
    n=read();size=min(n,6*(int)sqrt(n));
    for(register int i=1;i<=n;++i)s[i]=read(),++num2[s[i]];
    for(register int i=1;i<n;i+=size)
    {
        int r=min(i+size-1,n);
        for(register int j=i;j<=r;++j)--num2[s[j]];
        for(register int j=i;j<=r;++j)
        {
            for(int k=j+1;k<=r;k++)
            {
                int x=2*s[j]-s[k];
                if(x>=0)ans+=num1[x];
                x=2*s[k]-s[j];
                if(x>=0)ans+=num2[x];
            }
            ++num1[s[j]];
        }
    }
    for(register int i=1;i<n;i+=size)
    {
        int r=min(i+size-1,n),mx=0;
        for(register int j=1;j<i;++j)a[s[j]].r++,mx=max(mx,s[j]);
        for(register int j=r+1;j<=n;++j)b[s[j]].r++,mx=max(mx,s[j]);
        for(N=1;N<=mx;N<<=1);N<<=1;
        init(N);fft(a,0);fft(b,0);
        for(register int j=0;j<N;++j)a[j]=a[j]*b[j];
        fft(a,1);
        for(register int j=i;j<=r;++j) ans+=(ll)(a[s[j]<<1].r+0.5);
        memset(a,0,sizeof(cp)*N);
        memset(b,0,sizeof(cp)*N);
    }
    printf("%lld
",ans);
    return 0;
}