标签：ber line tree nsis sam 情况 rest free otherwise
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.
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 \(N(3≤N≤10 ^5)\).
The following line contains N integers \(a_i (1≤a_i≤10 ^ 5)\), which denotes the length of each branch, respectively.
Output the probability that their branches can form a triangle, in accuracy of 7 decimal places.
2
4
1 3 3 4
4
2 3 3 40.5000000
1.0000000从n根木棍取出三根，问组成三角形的概率有多大
首先取出的全部方案即为\(C_{n}^3\),我们只需要算出合法的方案数,两数相除即为概率
合法的方案考虑3种情况
一是选择一根长度为c的,然后选择两根长度小于c,且和大于c的,长度小于c的木棍数\(sum1\)我们可以直接前缀和求,我们用fft可以求得两个木棍长度之和小于等于c的\(sum2\),然后\(sum1*(sum1-1)/2-sum2\)即为两根长度小于c的组成的和大于c的方案数,若长度为c的木棍共有\(cnt[c]\)根,则此时方案数即为\(cnt[c]*(sum1*(sum1-1)/2-sum2)\)
二是选择两根长度为c的,方案数为\(cnt[c]*(cnt[c]-1)/2\),然后选择一根长度小于c的,方案数为\(cnt[c]*(cnt[c]-1)/2*sum1\)
三是选择三根长度为c的,方案数为\(C_{cnt[c]}^3\)
for一遍全都加起来即可
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
typedef long long ll;
struct cp {
double r, i;
cp(double r = 0, double i = 0): r(r), i(i) {}
cp operator + (const cp &b) {
return cp(r + b.r, i + b.i);
}
cp operator - (const cp &b) {
return cp(r - b.r, i - b.i);
}
cp operator * (const cp &b) {
return cp(r * b.r - i * b.i, r * b.i + i * b.r);
}
};
void change(cp a[], int len) {
for (int i = 1, j = len / 2; i < len - 1; i++) {
if (i < j) swap(a[i], a[j]);
int k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if (j < k) j += k;
}
}
void fft(cp a[], int len, int op) {
change(a, len);
for (int h = 2; h <= len; h <<= 1) {
cp wn(cos(-op * 2 * pi / h), sin(-op * 2 * pi / h));
for (int j = 0; j < len; j += h) {
cp w(1, 0);
for (int k = j; k < j + h / 2; k++) {
cp u = a[k];
cp t = w * a[k + h / 2];
a[k] = u + t;
a[k + h / 2] = u - t;
w = w * wn;
}
}
}
if (op == -1) {
for (int i = 0; i < len; i++) {
a[i].r /= len;
}
}
}
const int N = 8e5 + 50;
cp a[N];
ll b[N];
ll cnt[N];
ll sum1[N];
ll ans[N];
ll sum2[N];
int main() {
int t; scanf("%d", &t);
while (t--) {
for (int i = 0; i <= 4e5; i++) {
a[i] = cp(0, 0);
b[i] = 0;
cnt[i] = 0;
sum1[i] = 0;
sum2[i] = 0;
ans[i] = 0;
}
int n;
scanf("%d", &n);
int len1 = 0;
for (int i = 0; i < n; i++) {
int x;
scanf("%d", &x);
len1 = max(len1, x + 1);
cnt[x]++;
a[x].r++;
b[x + x]++;
}
int len = 1;
while (len < 2 * len1) len <<= 1;
fft(a, len, 1);
for (int i = 0; i < len; i++) {
a[i] = a[i] * a[i];
}
fft(a, len, -1);
for (int i = 0; i < len; i++) {
ans[i] = (ll)(a[i].r + 0.5);
ans[i] -= b[i];
ans[i] /= 2;
}
// for (int i = 0; i < 10; i++) {
// printf("%d : %d\n", i, ans[i]);
// }
for (int i = 0; i < len; i++) {
if (i) {
sum1[i] = sum1[i - 1] + cnt[i];
sum2[i] = sum2[i - 1] + ans[i];
}
else {
sum1[i] = cnt[i];
sum2[i] = ans[i];
}
}
ll tot = (ll)n * (n - 1) * (n - 2) / 6;
ll res = 0;
for (int i = 1; i < len; i++) {
if (cnt[i] >= 1) res += cnt[i] * (sum1[i - 1] * (sum1[i - 1] - 1) / 2 - sum2[i]);
if (cnt[i] >= 2) res += cnt[i] * (cnt[i] - 1) / 2 * sum1[i - 1];
if (cnt[i] >= 3) res += cnt[i] * (cnt[i] - 1) * (cnt[i] - 2) / 6;
}
//printf("%lld %lld\n", res, tot);
printf("%.7f\n", (double)(res) / (double)tot);
}
return 0;
}标签：ber line tree nsis sam 情况 rest free otherwise
原文地址：https://www.cnblogs.com/artoriax/p/12187434.html
