3-idiots
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2632 Accepted Submission(s): 900
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.
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.
用FFT算出二元和,枚舉最長邊,判重
/** Author: ☆·aosaki(*’(OO)’*) niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
return a==0?b:gcd(b%a,a);
}
#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 400044
using namespace std;
struct complex
{
double r,i;
complex(double _r = 0,double _i = 0)
{
r = _r; i = _i;
}
complex operator +(const complex &b)
{
return complex(r+b.r,i+b.i);
}
complex operator -(const complex &b)
{
return complex(r-b.r,i-b.i);
}
complex operator *(const complex &b)
{
return complex(r*b.r-i*b.i,r*b.i+i*b.r);
}
};
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].r/=len;
}
complex x1[MAX];
int a[MAX/4];
ll num[MAX];//超int!!!
ll sum[MAX];
int main()
{
//freopen("in.txt","r",stdin);
int T,n;
sc(T);
while(T--)
{
sc(n);
mem(num);
lp0(i,n)
{
sc(a[i]);
num[a[i]]++;
}
sort(a,a+n);
int len1=a[n-1]+1;
int len=1;
while(len<2*len1) len<<=1; //補長
lp0(i,len1)
x1[i]=complex(num[i],0);
lpn(i,len1,len-1)
x1[i]=complex(0,0); //補0
fft(x1,len,1);
lp0(i,len)
x1[i] = x1[i]*x1[i];
fft(x1,len,-1);
for(int i=0;i<len;i++)
num[i]=(long long)(x1[i].r+0.5); //四捨五入
len=2*a[n-1]; //最大值
lp0(i,n) num[a[i]+a[i]]--; //兩個相同
lp(i,len)
{
num[i]/=2; //無序,除以2
}
sum[0]=0;
for(int i=1;i<=len;i++)
sum[i]=sum[i-1]+num[i];
ll cnt=0;
lp0(i,n)
{
cnt += sum[len]-sum[a[i]];
cnt -= (ll)(n-1-i)*i;//減掉一個取大,一個取小的
cnt -= (n-1);//減掉一個取本身,另外一個取其它
cnt -= (ll)(n-1-i)*(n-i-2)/2;//減掉大於它的取兩個的組合
}
ll re=(ll)n*(n-1)*(n-2)/6; //總數
printf("%.7lf\n",(double)cnt/re);
}
return 0;
}