HDU 4609 3-idiots (组合数学 + FFT)

题意：给定 n 条边，问随机选出 3 条边，能组成三角形的概率是多少。

析：答案很明显就是  能组成三角形的种数 / (C(n, 3))。现在的问题是怎么求能组成三角形的种数。

这个博客说的非常清楚了。。。

https://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html

总体来说就是把边长转换成下标，然后再根据组合数，就可以知道选出两条边，长度为 i 有多少种情况，然后再减去重复的，最后再枚举斜边，就可以解决这个问题了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 400000 + 100;
const int maxm = 1e6 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

struct Complex{
  double x, y;
  Complex(double x_ = 0., double y_ = 0.) : x(x_), y(y_) {}
  Complex operator - (const Complex &c) const{
    return Complex(x - c.x, y - c.y);
  }
  Complex operator + (const Complex &c) const{
    return Complex(x + c.x, y + c.y);
  }
  Complex operator * (const Complex &c) const{
    return Complex(x * c.x - y * c.y, x * c.y + c.x * y);
  }
};

void change(Complex *y, int len){
  for(int i = 1, j = (len>>1); i < len-1; ++i){
    if(i < j)  swap(y[i], y[j]);
    int k = len>>1;
    while(j >= k){
      j -= k;
      k >>= 1;
    }
    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(-1 == on)  for(int i = 0; i < len; ++i)
    y[i].x /= len;
}

int a[maxn>>2];
Complex x[maxn];
LL sum[maxn], num[maxn];

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);  ms(num, 0);
    for(int i = 0; i < n; ++i)  ++num[a[i]=readInt()];
    sort(a, a + n);
    int mmax = a[n-1] + 1;
    int len = 1;
    while(len < (mmax<<1))  len <<= 1;
    for(int i = 0; i < mmax; ++i)
      x[i] = Complex(num[i], 0);
    for(int i = mmax; i < len; ++i)
      x[i] = Complex();
    fft(x, len, 1);
    for(int i = 0; i < len; ++i)
      x[i] = x[i] * x[i];
    fft(x, len, -1);
    for(int i = 0; i < len; ++i)
      num[i] = (LL)(x[i].x + 0.5);
    for(int i = 0; i < n; ++i)
      --num[a[i]<<1];
    for(int i = 1; i <= len; ++i)
      sum[i] = sum[i-1] + num[i] / 2;
    LL ans = 0;
    for(int i = 0; i < n; ++i){
      ans += sum[len] - sum[a[i]];
      ans -= (LL)(n-i-1) * (n-i-2) / 2;  // both are greater than a[i]
      ans -= (LL)(n-i-1) * i;  // the one is greater than a[i] but the other is less;
      ans -= n - 1;  // the one of the branches is a[i]
    }
    LL de = (LL)n * (n-1) * (n-2) / 6;
    printf("%.7f\n", ans * 1. / de);
  }
  return 0;
}