HDU4609 3-idiots

Posted 2020-08-29 SilverNebula

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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≤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

 

 

Source

2013 Multi-University Training Contest 1

 

 

Recommend

liuyiding

 

卷积可以算出两个数组中各取一个数，可以得到某数值的方案数。

不能取两遍自己，且“取x再取y”和“取y再取x”等价，需要去重。

去重之后，枚举第三条边，减去不符合要求的方案数，最终得到符合要求的方案数。

↑FFT是用来优化求卷积过程的。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<complex>
 6 #include<cmath>
 7 #define pi acos(-1)
 8 #define LL long long
 9 using namespace std;
10 const int mxn=100010*3;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(ch>=‘0‘ && ch<=‘9‘){x=x*10-‘0‘+ch;ch=getchar();}
15     return x*f;
16 }
17 struct com{
18     double a,b;
19     com operator + (com &rhs){return (com){a+rhs.a,b+rhs.b};}
20     com operator - (com &rhs){return (com){a-rhs.a,b-rhs.b};}
21     com operator * (com &rhs){
22         return (com){a*rhs.a-b*rhs.b,a*rhs.b+b*rhs.a};
23     }
24 };
25 int n,m,L;
26 LL ans=0;
27 com a[mxn],b[mxn];
28 int rev[mxn];
29 void FFT(com *a,int flag){
30     int i,j;
31     for(i=0;i<n;i++)if(rev[i]>i)swap(a[i],a[rev[i]]);
32     for(i=1;i<n;i<<=1){
33         com wn=(com){cos(pi/i),flag*sin(pi/i)};
34         for(j=0;j<n;j+=(i<<1)){
35             com w=(com){1,0};
36             for(int k=0;k<i;k++,w=w*wn){
37                 com x=a[j+k],y=w*a[j+k+i];
38                 a[j+k]=x+y;
39                 a[j+k+i]=x-y;
40             }
41         }
42     }
43     if(flag==-1)for(i=0;i<n;i++) a[i].a/=n;
44     return;
45 }
46 int c[mxn];
47 LL cnt[mxn];
48 int main(){
49     int i,j;
50     int T=read();
51     
52     while(T--){
53         n=read();
54         for(i=0;i<n;i++) c[i]=read();
55         sort(c,c+n);
56         m=c[n-1]<<1;int tmp=n;L=0;
57         for(n=1;n<=m;n<<=1)L++;//
58         for(i=0;i<n;i++)
59             rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
60         memset(a,0,sizeof a);    memset(b,0,sizeof b);//
61         memset(cnt,0,sizeof cnt);
62         for(i=0;i<tmp;i++)cnt[c[i]]++;
63         for(i=0;i<=c[tmp-1];i++) a[i].a=cnt[i];
64         FFT(a,1);
65         for(i=0;i<n;i++) a[i]=a[i]*a[i];//自乘 
66         FFT(a,-1);
67         for(i=0;i<n;i++) cnt[i]=(int)(a[i].a+0.5);
68         for(i=0;i<tmp;i++) cnt[c[i]+c[i]]--;
69         for(i=0;i<n;i++) cnt[i]/=2;
70         LL res=(LL)tmp*(tmp-1)*(tmp-2)/6;
71         ans=res;
72         for(int k=0,i=0;i<n;i++){
73             if(cnt[i]){
74                 while(k<tmp && c[k]<i)k++;
75                 if(k==tmp)break;
76                 ans-=cnt[i]*(tmp-k);
77             }
78         }
79         printf("%.7f\n",(double)ans/res);
80     }
81     return 0;
82 }