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  • hdu4609 3-idiots

    这里

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    using namespace std;
    typedef long long ll;
    int T, n, lim, c[100005], rev[270005], tmpcnt, a[100005];
    const double PI=acos(-1.0);
    ll d[270005];
    struct Complex{
    	double x, y;
    	Complex (double xx=0.0, double yy=0.0){
    		x = xx;
    		y = yy;
    	}
    	Complex operator+(const Complex &u)const{
    		return Complex(x+u.x, y+u.y);
    	}
    	Complex operator-(const Complex &u)const{
    		return Complex(x-u.x, y-u.y);
    	}
    	Complex operator*(const Complex &u)const{
    		return Complex(x*u.x-y*u.y, x*u.y+y*u.x);
    	}
    }aa[270005];
    void fft(Complex a[], int opt){
    	for(int i=0; i<lim; i++)
    		if(i<rev[i])
    			swap(a[i], a[rev[i]]);
    	for(int i=2; i<=lim; i<<=1){
    		Complex wn=Complex(cos(PI*2/i), opt*sin(PI*2/i));
    		int tmp=i>>1;
    		for(int j=0; j<lim; j+=i){
    			Complex w=Complex(1.0, 0.0);
    			for(int k=0; k<tmp; k++){
    				Complex tmp1=a[j+k], tmp2=w*a[j+k+tmp];
    				a[j+k] = tmp1 + tmp2;
    				a[j+k+tmp] = tmp1 - tmp2;
    				w = w * wn;
    			}
    		}
    	}
    	if(opt==-1)
    		for(int i=0; i<lim; i++)
    			a[i].x /= lim;
    }
    int main(){
    	cin>>T;
    	while(T--){
    		memset(c, 0, sizeof(c));
    		memset(aa, 0, sizeof(aa));
    		lim = 1;
    		tmpcnt = 0;
    		scanf("%d", &n);
    		for(int i=1; i<=n; i++){
    			scanf("%d", &a[i]);
    			c[a[i]]++;
    		}
    		sort(a+1, a+1+n);
    		while(lim<=a[n]+a[n])	lim <<= 1, tmpcnt++;
    		for(int i=0; i<lim; i++)
    			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(tmpcnt-1));
    		for(int i=1; i<=a[n]; i++)
    			aa[i].x = c[i];
    		fft(aa, 1);
    		for(int i=0; i<lim; i++)
    			aa[i] = aa[i] * aa[i];
    		fft(aa, -1);
    		for(int i=0; i<=a[n]+a[n]; i++)
    			d[i] =  (ll)(aa[i].x+0.5);
    		for(int i=1; i<=n; i++)
    			d[a[i]+a[i]]--;
    		for(int i=0; i<=a[n]+a[n]; i++)
    			d[i] /= 2;
    		for(int i=1; i<=a[n]+a[n]; i++)
    			d[i] += d[i-1];
    		ll ans=0;
    		for(int i=1; i<=n; i++){
    			ans += d[a[n]+a[n]] - d[a[i]];
    			ans -= (ll)(n-i)*(i-1);
    			ans -= (ll)(n-i)*(n-i-1)/2;
    			ans -= n - 1;
    		}
    		printf("%.7f
    ", (double)ans/((ll)n*(n-1)*(n-2)/6));
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8776668.html
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