HDU－4609－3-idiots

题目链接

题解

代码

#include <iostream>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;
/*
*  给出n条线段长度，问任取3根，组成三角形的概率
*  n ≤ 10^5 用FFT求可以组成三角形的取法有几种
*/
const double PI = acos(-1.0);

//  复数结构体
struct Complex
{
double x, y;    //  实部和虚部 x + yi
Complex(double _x = 0.0, double _y = 0.0)
{
x = _x;
y = _y;
}
Complex operator - (const Complex &b) const
{
return Complex(x - b.x, y - b.y);
}
Complex operator + (const Complex &b) const
{
return Complex(x + b.x, y + b.y);
}
Complex operator * (const Complex &b) const
{
return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
}
};

//  进行FFT和IFFT前的反转变换
//  位置i和（i二进制反转后的位置）互换
//  len必须去2的幂
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]);
}
//  交换护卫小标反转的元素，i < j保证交换一次
//  i做正常的+1，j左反转类型的+1，始终保持i和j是反转的
k = len / 2;
while (j >= k)
{
j -= k;
k /= 2;
}
if (j < k)
{
j += k;
}
}
return ;
}

//  FFT
//  len必须为2 ^ k形式
//  on == 1时是DFT，on == -1时是IDFT
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].x /= len;
}
}
}

const int MAXN = 400040;
Complex x1[MAXN];
int a[MAXN / 4];
long long num[MAXN];        //  100000 * 100000会超出int
long long sum[MAXN];

int main(int argc, const char * argv[])
{
int T;
int n;
cin >> T;
while (T--)
{
cin >> n;
memset(num, 0, sizeof(num));
for (int i = 0; i < n; i++)
{
cin >> a[i];
num[a[i]]++;
}
sort(a, a + n);     //  排序
int len1 = a[n - 1] + 1;
int len = 1;
while (len < 2 * len1)
{
len <<= 1;
}
for (int i = 0; i < len1; i++)
{
x1[i] = Complex(num[i], 0);
}
for (int i = len1; i < len; i++)
{
x1[i] = Complex(0, 0);
}
fft(x1, len, 1);
for (int i = 0; i < len; i++)
{
x1[i] = x1[i] * x1[i];
}
fft(x1, len, -1);
for (int i = 0; i < len; i++)
{
num[i] = (long long)(x1[i].x + 0.5);
}
len = 2 * a[n - 1];
//  减掉取两个相同的组合
for (int i = 0; i < n; i++)
{
num[a[i] + a[i]]--;
}
for (int i = 1; i <= len; i++)
{
num[i] /= 2;
}
sum[0] = 0;
for (int i = 1; i <= len; i++)
{
sum[i] = sum[i - 1] + num[i];
}
long long cnt = 0;
for (int i = 0; i < n; i++)
{
cnt += sum[len] - sum[a[i]];
//  减掉一个取大，一个取小
cnt -= (long long)(n - 1 - i) * i;
//  减掉一个取本身，另一个取其他
cnt -= (n - 1);
cnt -= (long long)(n - 1 - i) * (n - i - 2) / 2;
}
long long tot = (long long)n * (n - 1) * (n - 2) / 6;
printf("%.7lf\n", (double)cnt / tot);
}

return 0;
}

参考