HDU 4609 3-idiots

http://acm.hdu.edu.cn/showproblem.php?pid=4609

题意：给你一组数，问可以组成多少个三角形

分析：才知道原来有FFT这个算法。。。
         看书还没有看懂，暂且知道有这么个东西

          还是看牛人的题解http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html

#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;

}