3-idiots
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2382 Accepted Submission(s): 822
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
题意 ：给出n条边，问选出三条边能组成三角形的概率。
第一次搞FFT，了解了下卷积，具体实现是借鉴了别人的代码。
用num[i]表示长度为i的出现几次。对于样例1 3 3 4,我们得到num={0,1,0,2,1}，
num数组和num数组卷积的含义:就是从{1 3 3 4}取一个数，从{1 3 3 4}再取一个
数，他们的和每个值各有多少个？
即{0 1 0 2 1}*{0 1 0 2 1} 卷积的结果应该是{0 0 1 0 4 2 4 4 1 }。
这个结果的意义如下：
从{1 3 3 4}取一个数，从{1 3 3 4}再取一个数
取两个数和为 2 的取法是一种：1+1
和为 4 的取法有四种：1+3， 1+3 ，3+1 ，3+1
和为 5 的取法有两种：1+4 ，4+1；
和为 6的取法有四种：3+3,3+3,3+3,3+3,3+3
和为 7 的取法有四种： 3+4,3+4,4+3,4+3
和为 8 的取法有 一种：4+4
<span style="font-size:14px;">#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#define ll long long
using namespace std;
const double pi=acos(-1.0);
const int maxn=100010;
struct Complex
{
double a,b;
Complex(){}
Complex(double _a,double _b):a(_a),b(_b){}
Complex operator + (const Complex &p)
{
return Complex(a+p.a,b+p.b);
}
Complex operator - (const Complex &p)
{
return Complex(a-p.a,b-p.b);
}
Complex operator * (const Complex &p)
{
return Complex(a*p.a-b*p.b,a*p.b+b*p.a);
}
}s[maxn*4];
int n,len,cnt,a[maxn*4];
ll num[maxn*4],sum[maxn*4];
void initial()
{
len=1;
memset(num,0,sizeof(num));
memset(sum,0,sizeof(sum));
}
void input()
{
int co;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
num[a[i]]++;
}
}
void ready()
{
sort(a,a+n);
int Max=a[n-1]+1;
while(len<2*Max) len<<=1;
for(int i=0;i<Max;i++) s[i]=Complex(num[i],0);
for(int i=Max;i<len;i++) s[i]=Complex(0,0);
}
void change()
{
for(int i=1,j=len/2;i<len-1;i++)
{
if(i<j) swap(s[i],s[j]);
int k=len/2;
while(j>=k)
{
j-=k;
k/=2;
}
if(j<k) j+=k;
}
}
void FFT(int on)
{
change();
for(int i=2;i<=len;i<<=1)
{
Complex wn=Complex(cos(-on*2*pi/i),sin(-on*2*pi/i));
for(int j=0;j<len;j+=i)
{
Complex w(1,0);
for(int k=j;k<j+i/2;k++)
{
Complex u=s[k];
Complex t=w*s[k+i/2];
s[k]=u+t;
s[k+i/2]=u-t;
w=w*wn;
}
}
}
if(on==-1)
{
for(int i=0;i<len;i++)
s[i].a/=len;
}
}
void deal()
{
FFT(1);
for(int i=0;i<len;i++) s[i]=s[i]*s[i];
FFT(-1);
cnt=2*a[n-1];
for(int i=0;i<=cnt;i++) num[i]=(ll)(s[i].a+0.5);
}
void solve()
{
ll ans=0;
for(int i=0;i<n;i++) num[a[i]+a[i]]--; // 减去两次取相同的
for(int i=1;i<=cnt;i++) num[i]/=2; // 对于两次去不同的算了两边，去重
for(int i=1;i<=cnt;i++) sum[i]=sum[i-1]+num[i];
for(int i=0;i<n;i++)
{
ll t=sum[cnt]-sum[a[i]];
ll b=n-1; // 减去与a[i]组合的
ll c=(ll)i*(n-i-1); // 减去一大一小组合的
ll d=(ll)(n-i-1)*(n-i-2)/2; // 减去两大组合的
ans+=t-b-c-d;
}
ll mul=(ll)(n)*(n-1)*(n-2)/6;
printf("%.7lf\n",ans*1.0/mul);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
initial();
input();
ready();
deal();
solve();
}
return 0;
}</span>