2019牛客多校第二场BEddy Walker 2—

题意

从数字 $0$ 除法，每次向前走 $i$ 步，$i$ 是 $1 \sim K$ 中等概率随机的一个数，也就是说概率都是 $\frac{1}{K}$。求落在过数字 $N$ 额概率，$N=-1$ 表示无穷远。

分析

设落在过 $i$ 的概率为 $p_i$，则 $p_i = \frac{1}{K}p_{i-1} + \frac{1}{K}p_{i-2}...+\frac{1}{K}p_{i-k}$.

以 $k=2$ 为例，

$p_0 = 1 \\
p_1 = \frac{1}{2} \\
p_2 = \frac{1}{2}(\frac{1}{2} + 1) = \frac{3}{4} \\
p_3 = \frac{1}{2}(\frac{3}{4} + \frac{1}{2}) = \frac{5}{8} \\
p_4 = \frac{1}{2}(\frac{5}{8} + \frac{3}{4}) = \frac{11}{16}$

容易推出 $p_n = \frac{\frac{2}{3}\cdot 2^n + \frac{1}{3}\cdot (-1)^n}{2^n}$,

可知，当 $n \to \infty$，$p_n=\frac{2}{3}$.

找规律，能发现 $n$ 为无穷大时 $p_n = \frac{2}{k+1}$

//$ 1 \leq K_i \leq 1021, -1 \leq N_i \leq 10^{18}$，矩阵快速幂会TLE的

#include <bits/stdc++.h>

using namespace std;

#define rep(i,a,n) for (long long i=a;i<n;i++)

#define per(i,a,n) for (long long i=n-1;i>=a;i--)

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define SZ(x) ((long long)(x).size())

typedef vector<long long> VI;

typedef long long ll;

typedef pair<long long,long long> PII;

const ll mod=;

ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

// head

long long _,n;

namespace linear_seq

{

    const long long N=;

    ll res[N],base[N],_c[N],_md[N];

    vector<long long> Md;

    void mul(ll *a,ll *b,long long k)

    {

        rep(i,,k+k) _c[i]=;

        rep(i,,k) if (a[i]) rep(j,,k)

            _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;

        for (long long i=k+k-;i>=k;i--) if (_c[i])

            rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;

        rep(i,,k) a[i]=_c[i];

    }

    long long solve(ll n,VI a,VI b)

    { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...

//        printf("%d\n",SZ(b));

        ll ans=,pnt=;

        long long k=SZ(a);

        assert(SZ(a)==SZ(b));

        rep(i,,k) _md[k--i]=-a[i];_md[k]=;

        Md.clear();

        rep(i,,k) if (_md[i]!=) Md.push_back(i);

        rep(i,,k) res[i]=base[i]=;

        res[]=;

        while ((1ll<<pnt)<=n) pnt++;

        for (long long p=pnt;p>=;p--)

        {

            mul(res,res,k);

            if ((n>>p)&)

            {

                for (long long i=k-;i>=;i--) res[i+]=res[i];res[]=;

                rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;

            }

        }

        rep(i,,k) ans=(ans+res[i]*b[i])%mod;

        if (ans<) ans+=mod;

        return ans;

    }

    VI BM(VI s)

    {

        VI C(,),B(,);

        long long L=,m=,b=;

        rep(n,,SZ(s))

        {

            ll d=;

            rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;

            if (d==) ++m;

            else if (*L<=n)

            {

                VI T=C;

                ll c=mod-d*powmod(b,mod-)%mod;

                while (SZ(C)<SZ(B)+m) C.pb();

                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;

                L=n+-L; B=T; b=d; m=;

            }

            else

            {

                ll c=mod-d*powmod(b,mod-)%mod;

                while (SZ(C)<SZ(B)+m) C.pb();

                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;

                ++m;

            }

        }

        return C;

    }

    long long gao(VI a,ll n)

    {

        VI c=BM(a);

        c.erase(c.begin());

        rep(i,,SZ(c)) c[i]=(mod-c[i])%mod;

        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));

    }

};

ll qpow(ll a, ll b, ll p)

{

    ll ret = ;

    while(b)

    {

        if(b & )  ret = ret * a %  p;

        a = a * a % p;

        b >>= ;

    }

    return ret;

}

int k;

vector<ll>p;

int main()

{

    int T;

    scanf("%d", &T);

    while(T--)

    {

        scanf("%d%lld", &k, &n);

        ll INV = qpow(k, mod-, mod);

        p.clear();

        p.push_back();

        for(int i = ;i < *k;i++)          //求出前2k项，给BM

        {

            ll tmp = ;

            for(int j = i-;j >= i-k && j >= ;j--)  tmp = (tmp + p[j]) % mod;

            p.push_back(tmp * INV % mod);

        }

        //for(int i = 0;i < 2*k;i++)  printf("%lld ", p[i]);

        //printf("\n");

        /*输出系数*/

        /*前k项递推，需要2*k项能确定*/

        //VI res = linear_seq::BM(p);

        //for(int i = 1;i < res.size();i++)  printf("%lld%c", (mod-res[i]) % mod, i == res.size()-1 ? '\n' : ' ');

        if(n == -)  printf("%lld\n", * qpow(k+, mod-, mod) % mod);

        else printf("%I64d\n",linear_seq::gao(p, n));

    }

}

参考链接：https://blog.nowcoder.net/n/c7beb081cf2247779d2fa198b73a6658

巴特西

2019牛客多校第二场BEddy Walker 2——BM递推

题意

分析

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