hdu 4609 （FFT求解三角形）

Problem Description

King OMeGa catched three men who had been streaking in the street. Looking as idiots though, the three men insisted that it was a kind of performance art, and begged the king to free them. Out of hatred to the real idiots, the king wanted to check if they were lying. The three men were sent to the king's forest, and each of them was asked to pick a branch one after another. If the three branches they bring back can form a triangle, their math ability would save them. Otherwise, they would be sent into jail.
However, the three men were exactly idiots, and what they would do is only to pick the branches randomly. Certainly, they couldn't pick the same branch - but the one with the same length as another is available. Given the lengths of all branches in the forest, determine the probability that they would be saved.

 

Input

An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.
Each test case begins with the number of branches N(3≤N≤10 ⁵).
The following line contains N integers a_i (1≤a_i≤10 ⁵), which denotes the length of each branch, respectively.

 

Output

Output the probability that their branches can form a triangle, in accuracy of 7 decimal places.

 

Sample Input

   
   
   
   

    
    
    
    2
4
1 3 3 4
4
2 3 3 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0.5000000
1.0000000
   
   
   
   

题意：给你n个边，求从其中选出3个组成三角形的概率

思路：参考-kuangbin大神

如果我们用num[i]表示长度为i的木棍有多少个，对于1 3 3 4就是

num[] = {0 1 0 2 1}从卷积的公式来看

乘法第k位置上的值 便是a[i]*b[j](i + j == k),如果位置表示的是长度，num[]表示的个数，那么卷积过后我们得到的便是两边和的个数

{0 1 0 2 1}*{0 1 0 2 1} 卷积的结果应该是{0 0  1  0  4  2  4  4  1 }。

在求出了两条边的和后，枚举第三边

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps=1e-10;
const int inf = 0x3f3f3f;
const int MOD = 1e9+7;

const double PI = acos(-1.0);

struct Complex
{
    double x,y;
    Complex(double _x = 0.0,double _y = 0.0)
    {
        x = _x;
        y = _y;
    }
    Complex operator-(const Complex &b)const
    {
        return Complex(x-b.x,y-b.y);
    }
    Complex operator+(const Complex &b)const
    {
        return Complex(x+b.x,y+b.y);
    }
    Complex operator*(const Complex &b)const
    {
        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
};

void change(Complex y[],int len)
{
    int i,j,k;
    for(i = 1,j = len/2; i < len-1; i++)
    {
        if(i < j) swap(y[i],y[j]);
        k = len/2;
        while(j >= k)
        {
            j-=k;
            k/=2;
        }
        if(j < k) j+=k;
    }
}

void fft(Complex y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j = 0; j < len; j+=h)
        {
            Complex w(1,0);
            for(int k = j; k < j+h/2; k++)
            {
                Complex u = y[k];
                Complex t = w*y[k+h/2];
                y[k] = u+ t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
    {
        for(int i = 0; i < len; i++)
            y[i].x /= len;
    }
}

const int maxn = 401000;
Complex x1[maxn],x2[maxn];
ll sum[maxn];
ll num[maxn];
int a[maxn];

int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int len = 1;
        int len1;
        memset(num,0,sizeof(num));
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&a[i]);
            num[a[i]]++;
        }
        sort(a,a+n);
        len1 = a[n-1] + 1;
        while(len < len1*2) len <<= 1;

        for(int i = 0; i < len1; i++)
            x1[i] = Complex(num[i],0);
        for(int i = len1; i < len; i++)
            x1[i] = Complex(0,0);

        fft(x1,len,1);
        //fft(x2,len,1);
        for(int i = 0; i < len; i++)
        {
            x1[i] =x1[i]*x1[i];
            //cout << x1[i].x << " "<< x1[i].y <<endl;
        }
        fft(x1,len,-1);
        for(int i = 0; i < len; i++)
        {
            sum[i] = (ll)(x1[i].x+0.5);
            //cout << sum[i] << endl;
        }
        len = a[n-1] * 2;
        for(int i = 0; i < n; i++) sum[a[i]+a[i]] --;
        for(int i = 1; i <= len; i++) sum[i] /= 2;
        for(int i = 1; i <= len; i++)
        {
            sum[i] += sum[i-1];
        }
        ll  tot = (ll)n*(n-1)*(n-2)/6;
        ll  ans = 0;

        for(int i = 0; i < n; i++)
        {
            ans += sum[len]-sum[a[i]];    //两边之和大于第三边

            ans -= (ll)(n-1-i) * i;      //一个比自己大，一个比自己小
            ans -= (n-1);                //取了自己
            ans -= (ll)(n-1-i)*(n-2-i)/2;    //都比自己大
        }
        //printf("%.7lf\n",(double)ans/tot);
        printf("%.7f\n",(double)ans/tot);
    }
    return 0;
}