题目链接
题意
一个喝醉的人每次步数是$1~k$中的等概率某一个数字，问他走到$n$的概率是多少。
思路
很容易想到dp,$dp[i]$表示走到$i$的概率。
这个线性递推显然不能用矩阵快速幂。
于是我学会了一个奇技淫巧，快速处理线性递推。1
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using namespace std;
typedef long long ll;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 1e5+ 5;
//#define rep(i,a,n) for (int i=a;i<n;i++)
//#define per(i,a,n) for (int 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) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1e9 + 7; //修改成题目要求的模数
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;}
ll Pow(ll a, ll b)
{
ll ans = 1;
a %= mod;
assert(b >= 0);
while (b)
{
if (b & 1)
ans = ans * a % mod;
b >>= 1;
a = a * a % mod;
}
return ans;
}
// head
namespace linear_seq
{
const int N = 10010;
ll res[N], base[N], _c[N], _md[N];
vector<ll> Md;
inline int SZ(vector<ll>& x)
{
return (int)x.size();
}
void mul(ll *a, ll *b, int k)
{
for (int i = 0; i < k + k; i++)
_c[i] = 0;
//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 (int i = 0; i < k; i++)
if (a[i])
for (int j = 0; j < k; j++)
_c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
for (int i = k + k - 1; i >= k; i--)
if (_c[i])
for (int j = 0; j < SZ(Md); j++)
_c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
for (int i = 0; i < k; i++)
a[i] = _c[i];
}
int solve(ll n, vector<ll> a, vector<ll> b)
{ // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
// printf("%d\n",SZ(b));
ll ans = 0, pnt = 0;
int k = SZ(a);
assert(SZ(a) == SZ(b));
//rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
for (int i = 0; i < k; i++)
_md[k - 1 - i] = -a[i];
_md[k] = 1;
Md.clear();
//rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
for (int i = 0; i < k; i++)
if (_md[i] != 0)
Md.push_back(i);
//rep(i,0,k) res[i]=base[i]=0;
for (int i = 0; i < k; i++)
res[i] = base[i] = 0;
res[0] = 1;
while ((1ll << pnt) <= n)
pnt++;
for (int p = pnt; p >= 0; p--)
{
mul(res, res, k);
if ((n >> p) & 1)
{
for (int 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;
for (int j = 0; j < SZ(Md); j++)
res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
}
}
//rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
for (int i = 0; i < k; i++)
ans = (ans + res[i] * b[i]) % mod;
if (ans < 0)
ans+=mod;
return ans;
}
vector<ll> BM(vector<ll> s)
{
vector<ll> C(1,1),B(1,1);
int L = 0, m = 1, b = 1;
for (int n = 0; n < SZ(s); n++)
{
ll d = 0;
for (int i = 0; i < L + 1; i++)
d = (d + (ll)C[i] * s[n-i]) % mod;
if (d==0)
++m;
else if (2 * L <= n)
{
vector<ll> T = C;
ll c = mod - d * Pow(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m)
C.push_back(0);
for (int i = 0; i < SZ(B); i++)
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 * Pow(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m)
C.push_back(0);
for (int i = 0; i < SZ(B); i++)
C[i + m] = (C[i + m] + c * B[i]) % mod;
++m;
}
}
return C;
}
ll gao(vector<ll> a,ll n)
{
vector<ll> c = BM(a);
c.erase(c.begin());
for (int i = 0; i < SZ(c); i++)
c[i] = (mod - c[i]) % mod;
dbg(SZ(c));
return solve(n, c, vector<ll> (a.begin(), a.begin() + SZ(c)));
}
};
vector<ll> v;
int main()
{
/*
int n;
while (~scanf("%d",&n))
{
vector<int>v; //输入前几项(一般2k项足够)
v.push_back(1);
v.push_back(1);
v.push_back(2);
v.push_back(3);
v.push_back(5);
v.push_back(8);
//VI{1,1,2,3,5,8} 解出斐波那契数列
printf("i:%d arr:%d\n",n,linear_seq::gao(v,n-1));
}
*/
int T;
scanf("%d", &T);
while (T--)
{
v.clear();
ll n;
int k;
scanf("%d%lld", &k, &n);
ll invk = Pow(k, mod - 2);
dbg(invk);
if (n < 0)
{
printf("%lld\n", 2 * Pow(k + 1, mod - 2) % mod);
continue;
}
v.push_back(1);
for (int i = 1; i < k; i++)
{
ll x = 0;
for (int j = 0; j < i; j++)
x = (x + v[j]) % mod;
x = x * invk % mod;
v.push_back(x);
}
for (int i = k; i <= 2 * k + 5; i++)
{
ll x = 0;
for (int j = 1; j <= k; j++)
x = (x + v[i - j]) % mod;
x = x * invk % mod;
v.push_back(x);
}
printf("%lld\n", linear_seq::gao(v, n));
}
return 0;
}