vjudge数学专题1

挑了几道可做的做了，难度大概升序排序

F New Year and Arbitrary Arrangement

New Year and Arbitrary Arrangement

期望dp，考虑逆推。

考虑设$f_{i,j}$为有$i$个$a$，当前序列有$j$个$ab$的期望长度。

有$f_{i,j}=f_{i+1,j}\times p_a+f_{i,j+i}\times p_b$

目标要设为为$f_{1,0}$，因为$f_{0,0}$会从自身转移，因为如果前面没有$a$的话会无限加$b$，目标状态设为$f_{1,0}$可以避免这个问题。

考虑边界，容易发现当$i+j\ge k$时就是边界，因为再加入一个$b$就停止了。

考虑如何求$f_{i,j}(i+j\ge k)$。

记$A=\frac{p_a}{p_a+p_b},B=\frac{p_b}{p_a+p_b}$

$$\begin{array}{rl}{f}_{i,j}& =B\times \sum _{a=0}^{\mathrm{\infty }}(i+j+a)A^a(\text{a是枚举又加入的}a\text{的个数})\\ & =B\times \sum _{i=0}^{\mathrm{\infty }}(c+i)A^i(\text{为方便，记c=i+j,将a替换为i})\\ & =(1-A)\times \sum _{i=0}^{\mathrm{\infty }}(c+i)A^i(\text{发现B=1-A，替换})\\ & =(1-A)\times \sum _{i=0}^n(c+i)A^i(\text{写}\mathrm{\infty }\text{不直观，将其替换为n，方便下面变形})\\ & =\sum _{i=0}^n(c+i)A^i-\sum _{i=0}^n(c+i)A^{i+1}(\text{拆开})\\ & =c+\sum _{i=1}^n(c+i)A^i-\sum _{i=1}^n(c+i-1)A^i+(c+n)\times A^{n+1}\\ & =c+\sum _{i=1}^n{A}^i+(c+n)\times A^{n+1}\\ & =i+j+\frac{p_a}{p_b}(\text{当}\underset{n\to \mathrm{\infty }}{lim},0<A<1\text{时，}(c+n)\times A^{n+1}=0\text{，}\sum _{i=1}^n{A}^i=\frac{A}{1-A}=\frac{p_a}{p_b})\end{array}$$

然后这道题用记搜就做完了。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define InF(x) freopen(x".in","r",stdin)
#define OutF(x) freopen(x".out","w",stdout)
#define ErrF(x) freopen(x".err","w",stderr)
#define AnsF(x) freopen(x".ans","w",stdout)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = InF("in"),*OutFile = OutF("out");
    // FILE *ErrFile = ErrF("err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 1e3 + 10,mod = 1e9 + 7;
inline int power(int a,int b,int mod){
    int res = 1;
    for(;b;b >>= 1,a = 1ll*a*a%mod) if(b&1) res = 1ll*res*a%mod;
    return res;
}
int k,pa,pb,inv,inv1,inv2,inv3,f[N][N];
bool vis[N][N];
int dp(int x,int y){
    if(x + y >= k) return (x+y+inv3)%mod;
    if(vis[x][y]) return f[x][y];
    vis[x][y] = true;
    int res = 0;
    res = (res + 1ll*dp(x+1,y)*inv1%mod)%mod;
    res = (res + 1ll*dp(x,x+y)*inv2%mod)%mod;
    return f[x][y] = res;
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    cin>>k>>pa>>pb;
    inv = power(pa+pb,mod-2,mod);
    inv1 = 1ll*pa*inv%mod,inv2 = 1ll*pb*inv%mod,inv3 = 1ll*pa*power(pb,mod-2,mod)%mod;
    cout<<dp(1,0);
}

E - Steps to One

由于个人习惯，以下的$n$表示题目中的$m$

设$f_i$为当前数列的$gcd$为$i$，且$gcd$变为1还需要的期望步数。

显然有$ans=1+\frac{\sum _{i=1}^n{f}_i}{n}$

转移方程为$f_k=1+\frac{\sum _{i=1}^n{f}_{gcd(i,k)}}{n}$

考虑优化$\sum _{i=1}^n{f}_{gcd(i,k)}$，推柿子。

看到了$gcd$。考虑莫反。

$$\begin{array}{rl}\sum _{i=1}^n{f}_{gcd(i,k)}& =\sum _{d|k}{f}_d\sum _{i=1}^n[gcd(i,k)=d]\\ & =\sum _{d|k}{f}_d\sum _{i=1}^{\frac{n}{d}}[gcd(i,\frac{k}{d})=1]\\ & =\sum _{d|k}{f}_d\sum _{i=1}^{\frac{n}{d}}\sum _{t|gcd(i,\frac{k}{d})}\mu (t)\\ & =\sum _{d|k}{f}_d\sum _{t|\frac{k}{d}}\mu (t)\lfloor \frac{n}{dt}\rfloor \\ & =\sum _{T|k}\lfloor \frac{n}{T}\rfloor \sum _{d|T}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )\end{array}$$

所以就有了$f_i=\frac{\sum _{T|k}\lfloor \frac{n}{T}\rfloor \sum _{d|T}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )}{n}+1$

再考虑把整个柿子当成一个以$f_i$为未知量的方程解。

$$\begin{array}{rl}{f}_i& =\frac{\sum _{T|k}\lfloor \frac{n}{T}\rfloor \sum _{d|T}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )}{n}+1\\ f_i& =\frac{\sum _{T|k}\lfloor \frac{n}{T}\rfloor \sum _{d|T,d<i}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )}{n}+1+\frac{\lfloor \frac{n}{i}\rfloor \times f_i}{n}\\ f_i& =\frac{\sum _{T|k}\lfloor \frac{n}{T}\rfloor \sum _{d|T,d<i}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )+n}{n-\lfloor \frac{n}{i}\rfloor }\end{array}$$

然后设$d{p}_i=\sum _{d|T,d<i}{f}_d\mu (\lfloor \frac{T}{d}\rfloor )$。

更新$dp$的话暴力跑就行，复杂度$O(n\log n)$。

至于$d<i$的条件，先更新$f$再更新$dp$就可以了。

复杂度$O(n\log n)$

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define InF(x) freopen(x".in","r",stdin)
#define OutF(x) freopen(x".out","w",stdout)
#define ErrF(x) freopen(x".err","w",stderr)
#define AnsF(x) freopen(x".ans","w",stdout)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = InF("in"),*OutFile = OutF("out");
    // FILE *ErrFile = ErrF("err");
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 1e5 + 10,mod = 1e9 + 7;
#define eb emplace_back
inline int power(int a,int b,int mod){
    int res = 1;
    for(;b;b >>= 1,a = 1ll*a*a%mod) if(b&1) res = 1ll*res*a%mod;
    return res;
}
int n,f[N],dp[N],mu[N],inv[N];
bitset<N> pd;
vector<int> prime,fac[N];
inline void get_mu(int n){
    mu[1] = 1;
    for(int i = 2;i <= n; ++i){
        if(!pd[i]) prime.eb(i),mu[i] = -1;
        for(int j:prime){
            if(i * j > n) break;
            pd[i*j] = true;
            if(i%j == 0) break;
            mu[i*j] = -mu[i];
        }
    }
}
inline void solve(){
    cin>>n;get_mu(n);
    rep(i,1,n,1) rep(j,i,n,i) fac[j].eb(i);
    rep(i,1,n,1) inv[i] = power(i,mod-2,mod);
    rep(i,1,n,1){
        f[i] = n;
        for(int T:fac[i]) f[i] = (f[i] + 1ll*(n/T)%mod*dp[T]%mod + mod)%mod;
        f[i] = 1ll*f[i]*power(n-n/i,mod-2,mod)%mod;
        for(int j = i;j <= n;j += i)
            dp[j] = (dp[j] + 1ll*f[i]*mu[j/i]%mod + mod)%mod;
    }
    int ans = 0;
    rep(i,1,n,1) ans = (ans + f[i])%mod;
    ans = 1ll * ans * inv[n]%mod;
    cout<<(ans+1)%mod; 
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}
 

持续更新中……