hdu6088 组合数+反演+拆系数fft

题意:两个人van石头剪子布的游戏一共n盘,假设A赢了a盘,B赢了b盘,那么得分是gcd(a,b),求得分的期望*(3^{2*n})
题解:根据题意很明显有(ans=3^{n}*sum_{a=0}^{n}sum_{b=0}^{n-a}gcd(a,b)C(n,a)C(n-a,b))
(ans=sum_{d=1}^ndsum_{a=0}^nsum_{b=0}^{n-a}[gcd(a,b)==d]C(n,a)C(n-a,b))
假设(f(d)=sum_{a=0}^nsum_{b=0}^{n-a}[gcd(a,b)==d]C(n,a)C(n-a,b)),(F(d)=sum_{a=0}^nsum_{b=0}^{n-a}[d|gcd(a,b)]C(n,a)C(n-a,b)),
那么(F(d)=sum_{d|x}f(x)),(f(d)=sum_{d|x}mu(frac{x}{d})F(x)).
(ans=3^{n}sum_{d=1}^ndsum_{d|x}mu(frac{x}{d})F(x))
(ans=3^{n}sum_{x=1}^nF(x)sum_{d|x}dmu(frac{x}{d}))
(ans=3^{n}sum_{x=1}^nF(x)phi(x))
(F(d)=sum_{a=0}^nsum_{b=0}^{n-a}[d|gcd(a,b)]C(n,a)C(n-a,b))
(F(d)=sum_{a=0}^{frac{n}{d}}sum_{b=0}^{frac{n}{d}-a}C(n,a*d)C(n-a*d,b*d)-1)
(F(d)=sum_{a=0}^{frac{n}{d}}sum_{b=0}^{frac{n}{d}-a}C(n,a*d+b*d)*C(a*d+b*d,b*d)-1)
(F(d)=sum_{a=0}^{frac{n}{d}}sum_{b=0}^{a}C(n,a*d)C(a*d,b*d)-1)
(F(d)=n!sum_{a=0}^{frac{n}{d}}frac{1}{(n-a*d)!}sum_{b=0}^{i}frac{1}{(j*d)!}*frac{1}{(i*d-j*d)!})
后面的求和用拆系数fft即可处理,枚举d计算F(d),复杂度(sum_{d=1}^nfrac{n}{d}log(frac{n}{d}))

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;
//using namespace __gnu_pbds;

const ld pi=acos(-1);
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=2000000+10,inf=0x3f3f3f3f;

struct cd{
    ld x,y;
    cd(ld _x=0.0,ld _y=0.0):x(_x),y(_y){}
    cd operator +(const cd &b)const{
        return cd(x+b.x,y+b.y);
    }
    cd operator -(const cd &b)const{
        return cd(x-b.x,y-b.y);
    }
    cd operator *(const cd &b)const{
        return cd(x*b.x - y*b.y,x*b.y + y*b.x);
    }
    cd operator /(const db &b)const{
        return cd(x/b,y/b);
    }
}a[N*3],b[N*3],dfta[N*3],dftb[N*3],dftc[N*3],dftd[N*3];
cd conj(cd a){return cd(a.x,-a.y);}
int rev[N*3],A[N],B[N],C[N*3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void fft(cd *a,int n,int dft)
{
    for(int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        cd wn(cos(dft*pi/step),sin(dft*pi/step));
        for(int j=0;j<n;j+=step<<1)
        {
            cd wnk(1,0);
            for(int k=j;k<j+step;k++)
            {
                cd x=a[k];
                cd y=wnk*a[k+step];
                a[k]=x+y;a[k+step]=x-y;
                wnk=wnk*wn;
            }
        }
    }
    if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;
}
void mtt(int n,int m,int p) {
    if(n<100&&m<100||min(n,m)<=5)
    {
        for(int i=0;i<=n+m;i++)C[i]=0;
        for(int i=0;i<=n;i++)for(int j=0;j<=m;j++)
        {
            C[i+j]+=1ll*A[i]*B[j]%p;
            if(C[i+j]>=p)C[i+j]-=p;
        }
        return ;
    }
    int sz=0;
    while((1<<sz)<=n+m)sz++;getrev(sz);
    int len=1<<sz;
    for(int i=0;i<len;i++)
    {
        int x=(i>n?0:A[i]%p),y=(i>m?0:B[i]%p);
        a[i]=cd(x&0x7fff,x>>15);
        b[i]=cd(y&0x7fff,y>>15);
    }
    fft(a,len,1);fft(b,len,1);
    for(int i=0;i<len;i++)
    {
        int j=(len-i)&(len-1);
        cd aa,bb,cc,dd;
        aa = (a[i] + conj(a[j])) * cd(0.5, 0);
        bb = (a[i] - conj(a[j])) * cd(0, -0.5);
        cc = (b[i] + conj(b[j])) * cd(0.5, 0);
        dd = (b[i] - conj(b[j])) * cd(0, -0.5);
        dfta[j] = aa * cc;dftb[j] = aa * dd;
        dftc[j] = bb * cc;dftd[j] = bb * dd;
    }
    for(int i=0;i<len;i++)
    {
        a[i] = dfta[i] + dftb[i] * cd(0, 1);
        b[i] = dftc[i] + dftd[i] * cd(0, 1);
    }
    fft(a,len,1);fft(b,len,1);
    for(int i=0;i<len;i++)
    {
        int da = (ll)(a[i].x / len + 0.5) % p;
        int bb = (ll)(a[i].y / len + 0.5) % p;
        int dc = (ll)(b[i].x / len + 0.5) % p;
        int dd = (ll)(b[i].y / len + 0.5) % p;
        C[i] = (da + ((ll)(bb + dc) << 15) + ((ll)dd << 30)) % p;
        C[i] = (C[i]+p)%p;
    }
}
int prime[N],cnt,phi[N];
bool mark[N];
void init()
{
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,phi[i]=i-1;
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
}
int n,p,fac,po;
int f(int d)
{
    int ans=0;
    for(int i=0;i<=n/d;i++)A[i]=prime[i*d],B[i]=prime[i*d];
    mtt(n/d,n/d,p);
//    for(int i=0;i<=n/d;i++)printf("%d ",C[i]);puts("");
    for(int i=0;i<=n/d;i++)
    {
        ans+=1ll*prime[n-i*d]*C[i]%p;
        if(ans>=p)ans-=p;
    }
    ans=(1ll*ans*fac-1+p)%p;
    return ans;
}
int main()
{
//    fin;
    init();
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&p);
        prime[0]=prime[1]=fac=po=1;
        for(int i=2;i<=n;i++)prime[i]=1ll*(p-p/i)*prime[p%i]%p;
        for(int i=1;i<=n;i++)prime[i]=1ll*prime[i-1]*prime[i]%p,fac=1ll*fac*i%p,po=1ll*po*3%p;
        int ans=0;
        for(int d=1;d<=n;d++)
        {
            ans+=1ll*phi[d]*f(d)%p;
            if(ans>=p)ans-=p;
        }
        printf("%d
",1ll*ans*po%p);
    }
    return 0;
}
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