HDU 4609 3-idiots FFT

3-idiots

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=4609

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≤105).
The following line contains N integers a_i (1≤a_i≤105), 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

Hint

题意

给你n个木棍，问你这些木棍我随便选3个，能够组成三角形的概率是多少

题解：

其实就是问能够组成三角形的种类有多少种

FFT求出所有A+B的种类之后，再暴力枚举最大值，然后容斥算出以这个为最大值的三角形有多少个

然后直接莽一波就完了

代码

#include<bits/stdc++.h>

using namespace std;

const int N = 600040;
const double pi = acos(-1.0);

int len;

struct Complex
{
    double r,i;
	Complex(double r=0,double i=0):r(r),i(i) {};
	Complex operator+(const Complex &rhs)
	{
		return Complex(r + rhs.r,i + rhs.i);
	}
	Complex operator-(const Complex &rhs)
	{
		return Complex(r - rhs.r,i - rhs.i);
	}
	Complex operator*(const Complex &rhs)
	{
		return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);
	}
} va[N],vb[N];

void rader(Complex F[],int len)	//len = 2^M,reverse F[i] with  F[j] j为i二进制反转
{
	int j = len >> 1;
	for(int i = 1;i < len - 1;++i)
	{
		if(i < j) swap(F[i],F[j]);	// reverse
		int k = len>>1;
		while(j>=k)
		{
			j -= k;
			k >>= 1;
		}
		if(j < k) j += k;
	}
}

void FFT(Complex F[],int len,int t)
{
	rader(F,len);
	for(int h=2;h<=len;h<<=1)
	{
		Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h));
		for(int j=0;j<len;j+=h)
		{
			Complex E(1,0);	//旋转因子
			for(int k=j;k<j+h/2;++k)
			{
				Complex u = F[k];
				Complex v = E*F[k+h/2];
				F[k] = u+v;
				F[k+h/2] = u-v;
				E=E*wn;
			}
		}
	}
	if(t==-1)	//IDFT
		for(int i=0;i<len;++i)
			F[i].r/=len;
}

void Conv(Complex a[],Complex b[],int len) //求卷积
{
	FFT(a,len,1);
	FFT(b,len,1);
	for(int i=0;i<len;++i) a[i] = a[i]*b[i];
	FFT(a,len,-1);
}
int n;
int a[N];
long long num[N],sum[N];
void solve()
{
    memset(num,0,sizeof(num));
    memset(sum,0,sizeof(sum));
    memset(va,0,sizeof(va));
    memset(vb,0,sizeof(vb));
    scanf("%d",&n);
    int Mx = 0;
    for(int i=0;i<n;i++)
    {
        int x;scanf("%d",&a[i]);
        Mx = max(Mx,a[i]);
        num[a[i]]++;
    }
    Mx*=2;
    len=1;
    while(len<=Mx+1)len*=2;
    sort(a,a+n);
    for(int i=0;i<=len;i++)
    {
        va[i].r=num[i];
        va[i].i=0;
        vb[i].r=va[i].r;
        vb[i].i=0;
    }
    Conv(va,vb,len);
    for(int i=0;i<len;i++)
        num[i]=(long long)(va[i].r+0.5);
    for(int i=0;i<n;i++)
        num[a[i]+a[i]]--;
    for(int i=0;i<len;i++)
        num[i]/=2LL;
    sum[0]=0;
    for(int i=1;i<=len;i++)
        sum[i]=sum[i-1]+num[i];
    long long cnt = 0;
    for(int i=0;i<n;i++)
    {
        cnt += sum[len]-sum[a[i]];
        cnt -= (long long)(n-1-i)*i;
        cnt -= (n-1);
        cnt -= (long long)(n-1-i)*(n-i-2)/2;
    }
    long long all = (long long)n*(n-1)*(n-2)/6;
    printf("%.7f
",(double)cnt/all);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
	return 0;
}