HDU 4609 3-idiots ——FFT

【题目分析】

一堆小木棍，问取出三根能组成三角形的概率是多少。

Kuangbin的博客中讲的很详细。

构造一个多项式 ai=i的个数。

然后卷积之后去重。

统计也需要去重。

挺麻烦的一道题。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>

#include <map>
#include <set>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;

#define maxn 500005
#define db double
#define ll long long
#define inf 0x3f3f3f3f
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)

void Finout()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    #endif
}

int Getint()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}

struct Complex{
	double x,y;
	Complex operator + (Complex a) { Complex ret; ret.x=x+a.x; ret.y=y+a.y; return ret;};
	Complex operator - (Complex a) { Complex ret; ret.x=x-a.x; ret.y=y-a.y; return ret;};
	Complex operator * (Complex a) { Complex ret; ret.x=x*a.x-y*a.y; ret.y=x*a.y+y*a.x; return ret;};
}a[maxn];

ll rev[maxn],n,m,len,T,b[maxn],sum;
ll pre_sum[maxn],cnt;
const double pi=acos(-1.0);

void FFT(Complex * x,int n,int f)
{
	F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]);
	for (int m=2;m<=n;m<<=1)
	{
		Complex wn;
		wn.x=cos(2.0*pi/m*f); wn.y=sin(2.0*pi/m*f);
		for (int i=0;i<n;i+=m)
		{
			Complex w;
			w.x=1; w.y=0;
			for (int j=0;j<(m>>1);++j)
			{
				Complex u=x[i+j],v=x[i+j+(m>>1)]*w;
				x[i+j]=u+v; x[i+j+(m>>1)]=u-v;
				w=w*wn;
			}
		}
	}
}

bool cmp(int a,int b){return a<b;}

int main()
{
    Finout();T=Getint();
    while (T--)
    {
    	memset(a,0,sizeof a);
    	cnt=0;
    	sum=n=Getint();
    	F(i,1,n) b[i]=Getint(),a[b[i]].x+=1;
		sort(b+1,b+sum+1,cmp);
		m=1,len=0;n=b[sum]*2+1;
		while (m<=n) m<<=1,len++; n=m;
		F(i,0,n-1)
		{
			int t=i,r=0;
			F(j,1,len) r<<=1,r|=t&1,t>>=1;
			rev[i]=r;
		}
		FFT(a,n,1); F(i,0,n-1) a[i]=a[i]*a[i];
		FFT(a,n,-1);
		F(i,0,n-1) a[i].x=a[i].x/n;
		F(i,1,sum) a[b[i]<<1].x-=1;
		F(i,0,n-1) a[i].x/=2;
		pre_sum[0]=a[0].x+0.5;
		F(i,1,n-1) pre_sum[i]=pre_sum[i-1]+a[i].x+0.5;
		F(i,1,sum)
		{
			cnt+=pre_sum[n-1]-pre_sum[b[i]];
			cnt-=(ll)(i-1)*(sum-i);
			cnt-=(ll)(sum-1);
			cnt-=(ll)(sum-i)*(sum-i-1)/2;
		}
		ll tot=(ll)sum*(sum-1)*(sum-2)/6;
		printf("%.7f\n",(db)cnt/tot);
	}
}