HDU4609 NTT||FFT

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先用NTT求出两两组合的方案数
然后就能o(n) 求出能组成三角形的方案数
NTT:
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N=4e5+10;const ll mod=( 1ll << 47 ) * 7 * 4451 + 1 ;const ll g=3;ll mul( ll x, ll y ){    return ( x * y - ( long long ) ( x / ( long double ) mod * y + 1e-3 ) * mod + mod ) % mod ;}ll power ( ll a, ll b ){    ll res = 1, tmp = a ;    while ( b )    {        if ( b & 1 ) res = mul ( res, tmp ) ;        tmp = mul ( tmp, tmp ) ;        b >>= 1 ;    }    return res ;}void DFT ( ll y[], int n, bool rev ){    for ( int i = 1, j, t, k ; i < n ; ++ i )    {        for ( k = n >> 1, t = i, j = 0 ; k ; k >>= 1, t >>= 1 )        {            j = j << 1 | t & 1 ;        }        if ( i < j ) swap ( y[i], y[j] ) ;    }    for ( int s = 2, ds = 1 ; s <= n ; ds = s, s <<= 1 )    {        ll wn = power ( g, ( mod - 1 ) / s ) ;        if ( !rev ) wn = power ( wn, mod - 2 ) ;        for ( int k = 0 ; k < n ; k += s )        {            ll w = 1, t ;            for ( int i = k ; i < k + ds ; ++ i, w = mul ( w, wn ) )            {                y[i + ds] = ( y[i] - ( t = mul ( y[i + ds], w ) ) + mod ) % mod ;                y[i] = ( y[i] + t ) % mod ;            }        }    }}void NTT ( ll x1[], ll x2[], int n ){    DFT ( x1, n, 1 ) ;    DFT ( x2, n, 1 ) ;    for ( int i = 0 ; i < n ; ++ i ) x1[i] = mul ( x1[i], x2[i] ) ;    DFT ( x1, n, 0 ) ;    ll vn = power ( n, mod - 2 ) ;    for ( int i = 0 ; i < n ; ++ i ) x1[i] = mul ( x1[i], vn ) ;}ll s[N],vis[N],sum[N];ll x1[N],x2[N];int main(){    int t;    scanf("%d",&t);    while(t--)    {        ll n;        scanf("%lld",&n);        ll mx;        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%lld",&s[i]);            vis[s[i]]++;        }        sort(s,s+n);        mx=s[n-1]+1;        int len=1;        while(len<2*mx) len<<=1;        for(int i=0; i<mx; i++)            x2[i]=x1[i]=vis[i];        for(int i=mx; i<len; i++)            x2[i]=x1[i]=0;        NTT(x1,x2,len);        len=2*mx;        ll res=0;        for(int i=0;i<n;i++)            x1[s[i]+s[i]]--;        for(int i=1;i<=len;i++)            x1[i]>>=1;        for(int i=1;i<=len;i++)           sum[i]=sum[i-1]+x1[i];        for(int i=0;i<n;i++)        {            res+=sum[len]-sum[s[i]];            res-=(n-1);            res-=(n-1-i)*i;            res-=(n-1-i)*(n-i-2)/2;        }        printf("%.7lf\n",6.0*res/(n*(n-1)*(n-2)));    }}
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