2019牛客多校第二场BEddy Walker 2——BM递推

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=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head

long long _,n;
namespace linear_seq
{
    const long long N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<long long> Md;
    void mul(ll *a,ll *b,long long k)
    {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k)
            _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (long long i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,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=0,pnt=0;
        long long k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (long long p=pnt;p>=0;p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s)
    {
        VI C(1,1),B(1,1);
        long long L=0,m=1,b=1;
        rep(n,0,SZ(s))
        {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n)
            {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            }
            else
            {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,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,0,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 = 1;
    while(b)
    {
        if(b & 1)  ret = ret * a %  p;
        a = a * a % p;
        b >>= 1;
    }
    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-2, mod);
        p.clear();
        p.push_back(1);
        for(int i = 1;i < 2*k;i++)          //求出前2k项，给BM
        {
            ll tmp = 0;
            for(int j = i-1;j >= i-k && j >= 0;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 == -1)  printf("%lld\n",2 * qpow(k+1, mod-2, mod) % mod);
        else printf("%I64d\n",linear_seq::gao(p, n));
    }
}

 

 

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