HDU 4609 3-idiots

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4609
转载声明:http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html

题意:
现在我要跟着bin神的脚步来做题,,,虽然没人家聪明,我要一点一点理解fft,我会发现这样的题目,很显然很显然的让你措不及手,必然会想到这样的方法,但是呢,时间复杂度绝壁爆炸,, 而且对于数据量来说,都会设置很小,暂时这个情况来看…
我觉得最主要是理解 x^k 究竟代表什么. 这样做出来就行了,系数ak来表达->x^k

分析:fft

代码

/* Author:GavinjouElephant * Title: * Number: * main meanning： * * * */#include <iostream>using namespace std;#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <sstream>#include <cctype>#include <vector>#include <set>#include <cstdlib>#include <map>#include <queue>//#include<initializer_list>//#include <windows.h>//#include <fstream>//#include <conio.h>#define MaxN 0x7fffffff#define MinN -0x7fffffff#define lson 2*k#define rson 2*k+1typedef long long ll;const int INF=0x3f3f3f3f;const int maxn=1e5+10;int Scan()//读入整数外挂.{    int res = 0, ch, flag = 0;    if((ch = getchar()) == '-')             //判断正负        flag = 1;    else if(ch >= '0' && ch <= '9')           //得到完整的数        res = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9' )        res = res * 10 + ch - '0';    return flag ? -res : res;} void Out(int a)    //输出外挂 {     if(a>9)         Out(a/10);     putchar(a%10+'0'); } 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; } Complex x1[maxn*4]; int a[maxn]; long long num[maxn*4]; long long sum[maxn*4]; int T; int n;int main(){#ifndef ONLINE_JUDGE    freopen("coco.txt","r",stdin);    freopen("lala.txt","w",stdout);#endif    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];        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/tot);    }    return 0;}