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\(ans=3^{n}*\sum_{a=0}^{n}\sum_{b=0}^{n-a}gcd(a,b)C(n,a)C(n-a,b)\)

\(ans=\sum_{d=1}^nd\sum_{a=0}^n\sum_{b=0}^{n-a}[gcd(a,b)==d]C(n,a)C(n-a,b)\)

假设\(f(d)=\sum_{a=0}^n\sum_{b=0}^{n-a}[gcd(a,b)==d]C(n,a)C(n-a,b)\),\(F(d)=\sum_{a=0}^n\sum_{b=0}^{n-a}[d|gcd(a,b)]C(n,a)C(n-a,b)\),

那么\(F(d)=\sum_{d|x}f(x)\),\(f(d)=\sum_{d|x}\mu(\frac{x}{d})F(x)\).

\(ans=3^{n}\sum_{d=1}^nd\sum_{d|x}\mu(\frac{x}{d})F(x)\)

\(ans=3^{n}\sum_{x=1}^nF(x)\sum_{d|x}d\mu(\frac{x}{d})\)

\(ans=3^{n}\sum_{x=1}^nF(x)\phi(x)\)

\(F(d)=\sum_{a=0}^n\sum_{b=0}^{n-a}[d|gcd(a,b)]C(n,a)C(n-a,b)\)

\(F(d)=\sum_{a=0}^{\frac{n}{d}}\sum_{b=0}^{\frac{n}{d}-a}C(n,a*d)C(n-a*d,b*d)-1\)

\(F(d)=\sum_{a=0}^{\frac{n}{d}}\sum_{b=0}^{\frac{n}{d}-a}C(n,a*d+b*d)*C(a*d+b*d,b*d)-1\)

\(F(d)=\sum_{a=0}^{\frac{n}{d}}\sum_{b=0}^{a}C(n,a*d)C(a*d,b*d)-1\)

\(F(d)=n!\sum_{a=0}^{\frac{n}{d}}\frac{1}{(n-a*d)!}\sum_{b=0}^{i}\frac{1}{(j*d)!}*\frac{1}{(i*d-j*d)!}\)

后面的求和用拆系数fft即可处理,枚举d计算F(d),复杂度\(\sum_{d=1}^n\frac{n}{d}log(\frac{n}{d})\)

//#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\n",1ll*ans*po%p);

    }

    return 0;

}

/********************

********************/

巴特西

hdu6088 组合数+反演+拆系数fft

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