HDU－4609－3-idiots

ACM模版

题目链接

3-idiots

题解

给出n条线段长度，任取三根，组成三角形的概率。
使用FFT求可以组成三角形的取法有几种。

代码

#include 
#include 
#include 
#include 
#include 

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

《FFT》