【HDU 4609】3-idiots

1.题目链接。题目大意：给出n根棍子，现在从中任意选三种，能够组成三角形的概率。

2.其实就是说在这些棍子能够组成多少种三角形。我们预处理a+b,然后枚举第三条边c.预处理a+b采用FFT加速即可。然后统计答案的时候，注意把不合法的减了。

#include<bits/stdc++.h>
using namespace std;
#pragma warning(disable:4996)
const double PI = acos(-1.0);
struct Complex
{double r, i;Complex(double _r = 0, double _i = 0){r = _r; i = _i;}Complex operator +(const Complex &b){return Complex(r + b.r, i + b.i);}Complex operator -(const Complex &b){return Complex(r - b.r, i - b.i);}Complex operator *(const Complex &b){return Complex(r*b.r - i * b.i, r*b.i + i * b.r);}
};
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]);k = len / 2;while (j >= k){j -= k;k /= 2;}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 (on == -1)for (int i = 0; i < len; i++)y[i].r /= len;
}const int MAXN = 400040;
Complex x1[MAXN];
int a[MAXN / 4];
long long num[MAXN];//100000*100000会超int
long long sum[MAXN];int main()
{int T;int n;scanf("%d", &T);while (T--){scanf("%d", &n);memset(num, 0, sizeof(num));for (int i = 0; i < n; i++){scanf("%d", &a[i]);num[a[i]]++;}sort(a, a + n);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].r + 0.5);len = 2 * a[n - 1];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]];cnt -= (long long)(n - 1 - i)*i;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;
}

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