Hdu 4609 (FFT)

题目链接

3-idiots

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2165    Accepted Submission(s): 740

Problem Description

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.

Input

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≤10⁵).
The following line contains N integers a_i (1≤a_i≤10⁵), which denotes the length of each branch, respectively.

Output

Output the probability that their branches can form a triangle, in accuracy of 7 decimal places.

Sample Input

2 4 1 3 3 4 4 2 3 3 4

Sample Output

0.5000000 1.0000000

给出ｎ个数，从中选出三个数，问能组成三角形的概率。

首先记录每个数出现的次数。然后和自己做多项式乘法，得到两个数组合起来可以拼成的数的情况。

当然，这里应该去掉自身和自身组合，也要考虑组合的无序性。

然后将所有数排序，假设a[i]作为三角形中最大的边。代码里有相关注释。

Accepted Code:

#define _CRT_SECURE_NO_WARNINGS
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <bitset>
#include <complex>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii;
typedef long long ll; typedef vector<long long> vl; typedef pair<long long,long long> pll; typedef vector<pair<long long,long long> > vpll;
typedef vector<string> vs; typedef long double ld;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

typedef long double Num;    //??????long double?????
const Num PI = 3.141592653589793238462643383279L;
typedef complex<Num> Complex;
//n?????
//a?????
void fft_main(int n, Num theta, Complex a[]) {
    for(int m = n; m >= 2; m >>= 1) {
        int mh = m >> 1;
        Complex thetaI = Complex(0, theta);
        rep(i, mh) {
            Complex w = exp((Num)i*thetaI);
            for(int j = i; j < n; j += m) {
                int k = j + mh;
                Complex x = a[j] - a[k];
                a[j] += a[k];
                a[k] = w * x;
            }
        }
        theta *= 2;
    }
    int i = 0;
    reu(j, 1, n-1) {
        for(int k = n >> 1; k > (i ^= k); k >>= 1) ;
        if(j < i) swap(a[i], a[j]);
    }
}

void fft(int n, Complex a[]) { fft_main(n, 2 * PI / n, a); }
void inverse_fft(int n, Complex a[]) { fft_main(n, -2 * PI / n, a); }

void convolution(vector<Complex> &v, vector<Complex> &w) {
    int n = 1, vwn = v.size() + w.size() - 1;
    while(n < vwn) n <<= 1;
    v.resize(n), w.resize(n);
    fft(n, &v[0]);
    fft(n, &w[0]);
    rep(i, n) v[i] *= w[i];
    inverse_fft(n, &v[0]);
    rep(i, n) v[i] /= n;
}

//　solve problem....
const int MAXN = 100002;
int a[MAXN];
long long num[MAXN], sum[MAXN<<2];

void read(int &res) {
    res = 0;
    char c = ' ';
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
}

vector<long long> calc(const vector<long long> &A) {
    vector<Complex> Lc(all(A)), Rc(all(A));
    convolution(Lc, Rc);
    long long n = A.size() + A.size() - 1;
    vector<long long> res(n);
    rep(i, n) res[i] = (long long)(Lc[i].real() + .5);
//  rep(i, n) cerr << res[i] << ", "; cerr << endl;
    return res;
}

int main() {
    int T; read(T);
    while (T--) {
        int N; read(N);
        //a[i]出现次数
        memset(num, 0, sizeof(num));
        rep(i, N) {
            read(a[i]);
            num[a[i]]++;
        }
        sort(a, a + N);
        int len = a[N-1] * 2;
        vector<long long> A(num, num + a[N-1] + 1);
        vector<long long> ans = calc(A);
        //重复使用
        rep(i, N) {
            ans[a[i]+a[i]]--;
        }
        //无序
        rep(i, len) {
            ans[i+1] /= 2;
        }
        //前缀和
        sum[0] = 0;
        rep(i, len) {
            sum[i+1] = sum[i] + ans[i+1];
        }
        long long cnt = 0;
        //假设a[i] 为最大的边
        rep(i, N) {
            cnt += sum[len] - sum[a[i]];
            //a[i]本身
            cnt -= N - 1;
            //一大一小
            cnt -= (long long)i * (N - i - 1); 
            //两大
            cnt -= (long long)(N-i-1) * (N-i-2) / 2;
        }
        //总的方法数 = C(N, 3)
        long long tot = (long long)N * (N-1) * (N-2) / 6;
        printf("%.7f
", (cnt+0.0) / tot);
    }
    return 0;
}