快速傅里叶变换FFT——kuangbin模板

/*
#include <bits/stdc++.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("%.7f\n",(double)cnt/tot);
    }
    return 0;
}*/

#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int MAXN = 400044;
struct Complex{
    double x,y;
    Complex(double _x = 0.0,double _y =0.0){
        x = _x;y =_y;
    }
    Complex operator - (const Complex &b)const{
        return Complex(x-b.x,y-b.y);
    }
    Complex operator + (const Complex &b)const{
        return Complex(x+b.x,y+b.y);
    }
    Complex operator * (const Complex &b)const{
        return Complex(x*b.x - y*b.y,x*b.y + y*b.x);
    }
};
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].x /= len;

}

Complex x1[MAXN];
int a[MAXN/4];
long long num[MAXN];
long long sum[MAXN];
int main()
{
    int T;
    cin >> T;
    int n;
    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].x + 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("%.7f\n",(double)(cnt*1.0/tot));
    }
    return 0;
}

 

posted @ 2017-08-11 13:41  HITLJR  阅读(1043)  评论(0编辑  收藏  举报