HDU 4609 3-idiots(FFT)

给你n个木棍，长度都是10W以内，问你选三根构成三角形的概率
数据范围小的话应该有各种n2的姿势
但是现在给10W，考虑母函数，长度作为指数，系数是这个长度的个数
然后先考虑任选两根，能组合出的长度有多少种
两个相同的多项式做FFT，对于后来的长度，然后要去掉同一根选两次的
然后选1和2，和选2和1，是同一种，方案数还得除以2，然后求个系数的前缀和
然后现在有了所有两根木棍组合的方案数，考虑枚举最长的木棍是i
然后另外两根加起来要大于i，所以是sum[2∗maxn]−sum[i]
然后要去重啊，因为i是三根里最长的，可能一根选择的比他大，一根比他小，或者两根都比他大，或者其中一根选了这根自己

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代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           100010
#define   MAXN          1000005
#define   maxnode       5
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
//const double PI    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-6;
const LL     mod   = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
 }
/*****************************************************/
const double PI = acos(-1.0);
struct Complex  {  
    double real, image;  
    Complex(double _real, double _image)  {  
        real = _real;  
        image = _image;  
    }  
    Complex(){}  
};  

Complex operator + (const Complex &c1, const Complex &c2)  {  
    return Complex(c1.real + c2.real, c1.image + c2.image);  
}  

Complex operator - (const Complex &c1, const Complex &c2)  {  
    return Complex(c1.real - c2.real, c1.image - c2.image);  
}  

Complex operator * (const Complex &c1, const Complex &c2)  {  
    return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);  
}  

int rev(int id, int len)  {  
    int ret = 0;  
    for(int i = 0; (1 << i) < len; i++) {  
        ret <<= 1;  
        if(id & (1 << i)) ret |= 1;  
    }  
    return ret;  
}  

Complex A[280000];  /////////大小要修改，不要忘记
void FFT(Complex* a, int len, int DFT){//对a进行DFT或者逆DFT, 结果存在a当中    
    //Complex* A = new Complex[len]; 这么写会爆栈  
    for(int i = 0; i < len; i++)  
        A[rev(i, len)] = a[i];  
    for(int s = 1; (1 << s) <= len; s++)  {  
        int m = (1 << s);  
        Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));  
        for(int k = 0; k < len; k += m) {  
            Complex w = Complex(1, 0);  
            for(int j = 0; j < (m >> 1); j++) {  
                Complex t = w*A[k + j + (m >> 1)];  
                Complex u = A[k + j];  
                A[k + j] = u + t;  
                A[k + j + (m >> 1)] = u - t;  
                w = w*wm;  
            }  
        }  
    }  
    if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;  
    for(int i = 0; i < len; i++) a[i] = A[i];  
    return;  
}  

Complex a[280000], b[280000];//对应的乘积的长度不会超过100000, 也就是不超过(1 << 17) = 131072  
int num[MAX];
int x[MAX];
LL y[200005];
LL sum[200005];
int main(){
    //freopen("in.txt","r",stdin);
    int t;
    cin>>t; 
    while(t--){
        int n;
        cin>>n;
        mem(num,0);
        int maxn=0;
        for(int i=0;iscanf("%d",&x[i]);
            num[x[i]]++;
            maxn=max(maxn,x[i]);
        }
        int sa=0;
        while((1<1) sa++;
        int len=(1<<(sa+1));
        for(int i=0;iif(i<=maxn){
                a[i]=Complex(num[i],0);
                b[i]=Complex(num[i],0);
            }
            else{
                a[i]=Complex(0,0);
                b[i]=Complex(0,0);
            }
        }
        FFT(a,len,1);
        FFT(b,len,1);
        for(int i=0;i1);
        for(int i=0;i<=2*maxn;i++) y[i]=floor(a[i].real+0.5);
        for(int i=0;i<=2*maxn;i+=2) y[i]-=num[i/2];
        for(int i=0;i<=2*maxn;i++) y[i]/=2;
        sum[0]=y[0];
        for(int i=1;i<=2*maxn;i++) sum[i]=sum[i-1]+y[i];
        LL ans=0;
        sort(x,x+n);
        for(int i=0;i2*maxn]-sum[x[i]];
            ans-=(LL)(n-i-1)*i;
            ans-=(LL)(n-i-1)*(n-1-i-1)/2;
            ans-=n-1;
        }
        double ret=ans*6.0/n/(n-1)/(n-2);
        printf("%.7f\n",ret);
    }
    return 0;
}