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题目描述
题目大意:给定n个长度分别为ai的木棒,问随机选择3个木棒能够拼成三角形的概率。
题解
这道题要根据三角形的两边之和大于第三边来做,其实判定的条件就是随机选出来的三条边中较小的两条之和大于第三条边
首先容斥一下,用总的方案数减去不合法的方案数
令F(i)表示两条边和为i的方案数,a(i)表示长为i的木棒有多少个,那么可以写成一个卷积的形式
然后g(i)表示长度大于等于i的有多少个,那么不合法的数量就是
计算概率就行了
代码
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
#define N 300005
const double pi=acos(-1.0);
struct complex
{
double x,y;
complex(double X=0,double Y=0)
{
x=X,y=Y;
}
}a[N];
complex operator + (complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
complex operator - (complex a,complex b){return complex(a.x-b.x,a.y-b.y);}
complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
int T,cnt,Max,m,n,L,R[N];
int f[N],g[N];
LL up,down,F[N];
void clear()
{
cnt=Max=m=n=L=0;up=down=0;
memset(R,0,sizeof(R));memset(f,0,sizeof(f));memset(g,0,sizeof(g));memset(F,0,sizeof(F));
for (int i=0;i<=300000;++i) a[i]=complex(0,0);
}
void FFT(complex a[N],int opt)
{
for (int i=0;i<n;++i)
if (i<R[i]) swap(a[i],a[R[i]]);
for (int k=1;k<n;k<<=1)
{
complex wn=complex(cos(pi/k),opt*sin(pi/k));
for (int i=0;i<n;i+=(k<<1))
{
complex w=complex(1,0);
for (int j=0;j<k;++j,w=w*wn)
{
complex x=a[i+j],y=w*a[i+j+k];
a[i+j]=x+y,a[i+j+k]=x-y;
}
}
}
}
int main()
{
scanf("%d",&T);
while (T--)
{
clear();
scanf("%d",&cnt);
for (int i=1;i<=cnt;++i)
{
int x;scanf("%d",&x);
++a[x].x;++f[x];
Max=max(Max,x);
}
for (int i=Max;i>=1;--i) g[i]=g[i+1]+f[i];
m=Max<<1;
for (n=1;n<=m;n<<=1) ++L;
for (int i=0;i<n;++i)
R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
FFT(a,1);
for (int i=0;i<=n;++i) a[i]=a[i]*a[i];
FFT(a,-1);
for (int i=0;i<=n;++i)
{
F[i]=(LL)(a[i].x/n+0.5);
if (!(i&1))
F[i]-=(LL)f[i>>1];
F[i]>>=1;
}
up=down=(LL)cnt*(cnt-1)*(cnt-2)/6;
for (int i=0;i<=n;++i) up-=F[i]*(LL)g[i];
printf("%.7lf\n",(double)up/(double)down);
}
}