bzoj3513

给定n个长度分别为$a_i$的木棒，问随机选择3个木棒能够拼成三角形的概率。

$a_i$和$n$都在$10^5$以内

 

 

对于每一个i，我们统计比i短的边有多少组合能组成长度<=i的

用1减去这个概率就是能拼成的概率

具体就是用sum[i]表示i出现的次数

sum[i]可以转化成如下卷积的样子

$$sum[i] = \sum_{j=1}^{i-1}sum[j] * sum[i - j - 1]$$

然后FFT

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 300010;
const double pi = acos(-1.0);
struct Cint
{
    double r,i;
    Cint() {r = i = 0.00;}
    Cint(double _r,double _i) : r(_r),i(_i){}
    Cint operator + (const Cint &b)const{return Cint(r + b.r,i + b.i);}
    Cint operator - (const Cint &b)const{return Cint(r - b.r,i - b.i);}
    Cint operator * (const Cint &b)const{return Cint(r * b.r - i * b.i,i * b.r + r * b.i);}
}s[maxn];
int a[maxn],LEN,n;
LL sum[maxn],ans;
inline void FFT_init(Cint *a,int len)
{
    for(int i = 1,j = len >> 1,k;i < len - 1;i++)
    {
        if(i < j)swap(a[i],a[j]);k = len;
        while(j >= (k>>=1)) j -= k;  
        if(j <= k) j += k;
    }
}
inline void FFT(Cint *a,int len,int f)
{
    FFT_init(a,len);int l,i,j,k;Cint u,v;
    for(l = 2;l <= len;l <<= 1)
    {
        i = l >> 1;
        Cint w(cos(-f * 2 * pi / l),sin(-f * 2 * pi / l));
        for(j = 0;j ^ len;j += l)
        {
            Cint wn(1.0,0.0);
            for(k = j;k ^ (i + j);k++)
            {
                u = a[k]; v = wn * a[i + k];
                a[k] = u + v;a[k + i] = u - v;
                wn = w * wn;
            }
        }
    }
    if(f == -1)
        for(i = 0;i < len;i++)a[i] . r /= len;
}

int main()
{
    //freopen("ou.txt","r",stdin);
    //freopen("x.txt","w",stdout);
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);int mx = 0;ans = 0;
        for(int i=0;i<=LEN + 1;i++)s[i] = Cint(0.0,0.0);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            mx = max(mx,a[i]);
        }
        for(LEN = 1;(LEN >> 1) < mx;LEN <<= 1);
        for(int i=1;i<=n;i++)s[a[i]].r += 1.0;
        FFT(s,LEN,1);
        for(int i=0;i<LEN;i++)s[i] = s[i] * s[i];
        FFT(s,LEN,-1);
        for(int i=1;i<=n;i++)s[a[i] * 2].r -= 1.0;
        for(int i=1;i<=LEN;i++)sum[i] = sum[i-1] + floor(s[i].r + 0.5);
        for(int i=1;i<=n;i++)ans += sum[a[i]];
        double pos = 3.0 * ans / n / (n - 1) / (n - 2);
        printf("%.7lf\n",1.0 - pos);
    }
}