HDU 4609 3-idiots [FFT 快速傅里叶变换]_给出三边范围求构成三角形的数量fft-优快云博客

本文链接：https://blog.youkuaiyun.com/Adolphrocs/article/details/101555137

探讨编程竞赛问题“3-Idiots”，任务是在给定的数组中找到能构成三角形的三个数的概率。使用FFT算法进行快速卷积运算，以高效解决此问题。

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3-idiots

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10046 Accepted Submission(s): 3406

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

Source

2013 Multi-University Training Contest 1

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题意：给出一个数组，求出取三个数能组成三角形的概率

题解：FFT bin巨题解https://www.cnblogs.com/kuangbin/p/3210565.html

首先题目给了a数组，

如样例一：

1 3 3 4

把这个数组转化成num数组，num[i]表示长度为i的有num[i]条。

样例一就是

num = {0 1 0 2 1}

代表长度0的有0根，长度为1的有1根，长度为2的有0根，长度为3的有两根，长度为4的有1根。

使用FFT解决的问题就是num数组和num数组卷积。

num数组和num数组卷积的解决，其实就是从{1 3 3 4}取一个数，从{1 3 3 4}再取一个数，他们的和每个值各有多少个

例如{0 1 0 2 1}*{0 1 0 2 1} 卷积的结果应该是{0 0 1 0 4 2 4 4 1 }

长度为n的数组和长度为m的数组卷积，结果是长度为n+m-1的数组。

{0 1 0 2 1}*{0 1 0 2 1} 卷积的结果应该是{0 0 1 0 4 2 4 4 1 }。

这个结果的意义如下：

从{1 3 3 4}取一个数，从{1 3 3 4}再取一个数

取两个数和为 2 的取法是一种：1+1

和为 4 的取法有四种：1+3， 1+3 ，3+1 ，3+1

和为 5 的取法有两种：1+4 ，4+1；

和为 6的取法有四种：3+3,3+3,3+3,3+3,3+3

和为 7 的取法有四种： 3+4,3+4,4+3,4+3

和为 8 的取法有一种：4+4

利用FFT可以快速求取循环卷积，具体求解过程不解释了，就是DFT和FFT的基本理论了。

总之FFT就是快速求到了num和num卷积的结果。只要长度满足>=n+m+1.那么就可以用循环卷积得到线性卷积了。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;

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];
        //减掉取两个相同的组合
        for(int i = 0;i < n;i++)
            num[a[i]+a[i]]--;
        //选择的无序，除以2
        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，一个取小的
            cnt -= (long long)(n-1-i)*i;
            //减掉一个取本身，另外一个取其它
            cnt -= (n-1);
            //减掉大于它的取两个的组合
            cnt -= (long long)(n-1-i)*(n-i-2)/2;
        }
        //总数
        long long tot = (long long)n*(n-1)*(n-2)/6;
        printf("%.7lf\n",(double)cnt/tot);
    }
    return 0;
}