HDU 4602 FFT 终于大致会用FFT了。。

还是去看他的题解吧。。好详细的说。。kuangbin大牛啊  

http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html

直接贴代码。。

/* *@author ipqhjjybj *@date 2013/08/12 */#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>const int N=1<<18;#define ll long longconst double pi=acos(-1.0);using namespace std;struct Complex{    double r,i;    Complex(double real=0.0,double image=0.0){        r=real;i=image;    }    Complex operator + (const Complex &o){        return Complex(r+o.r,i+o.i);    }    Complex operator - (const Complex &o){        return Complex(r-o.r,i-o.i);    }    Complex operator * (const Complex &o){        return Complex(r*o.r-i*o.i,r*o.i+i*o.r);    }}x1[N],x2[N];int a[N];ll sum[N],p[N];int l;void brc(Complex *y,int l){    register int i,j,k;    for(i=1,j=(l>>1);i<l-1;i++){        if(i<j) swap(y[i],y[j]);        k=l>>1;        while(j>=k){            j-=k;            k/=2;        }        if(j<k) j+=k;    }}void fft(Complex *y,int l,double on){    register int h,i,j,k;    Complex u,t;    brc(y,l);    for(h=2;h<=l;h<<=1){        Complex wn(cos(on*2*pi/h),sin(on*2*pi/h));        for(j=0;j<l;j+=h){            Complex w(1,0);            for(k=j;k<j+(h>>1);k++){                u=y[k];                t=w*y[k+(h>>1)];                y[k]=u+t;                y[k+(h>>1)]=u-t;                w=w*wn;            }        }    }    if(on==-1) for(i=0;i<l;i++)y[i].r/=l;}int main(void){    //freopen("4609.in","r",stdin);    register int i;    int t,n;    scanf("%d",&t);    while(t--){        l=1<<19;        scanf("%d",&n);        for(i=0;i<n;i++)            scanf("%d",a+i);        sort(a,a+n);        while((l>>2)>a[n-1]) l>>=1;        for(i=0;i<l;i++)            sum[i]=p[i]=0;        for(i=0;i<n;i++)            p[a[i]]++;        for(i=0;i<l;i++){            x2[i]=x1[i]=Complex(1.0*p[i],0.0);        }        fft(x1,l,1);        fft(x2,l,1);        for(i=0;i<l;i++)x1[i]=x1[i]*x2[i];        fft(x1,l,-1);        for(i=0;i<l;i++)sum[i]=(ll)(x1[i].r+0.5);        //for(i=0;i<l;i++){        //    printf("sum[%d]=%lld\n",i,sum[i]);        //}        for(i=0;i<n;i++)sum[a[i]+a[i]]--;        for(i=1;i<l;i++)sum[i]>>=1;        ll tot=(ll)n*(n-1)*(n-2)/6;        ll ans=tot;        int k=0;        for(i=0;i<l;i++)            if(sum[i]){                while(k<n&&a[k]<i)k++;                if(k==n)break;                ans-=sum[i]*(n-k);            }        printf("%.7lf\n",(double)ans/tot);    }    return 0;}