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

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

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$ans=1+\frac{\sum\limits_{i=1}^nf_i}{n}$

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

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

看到了$\gcd$。考虑莫反。

\[\begin{aligned} \sum_{i=1}^nf_{\gcd(i,k)}&=\sum_{d|k}f_d\sum_{i=1}^n[gcd(i,k)=d]\\ &=\sum_{d|k}f_d\sum_{i=1}^{\frac{n}{d}}[gcd(i,\frac{k}{d})=1]\\ &=\sum_{d|k}f_d\sum_{i=1}^{\frac{n}{d}}\sum_{t|gcd(i,\frac{k}{d})}\mu(t)\\ &=\sum_{d|k}f_d\sum_{t|\frac{k}{d}}\mu(t)\left\lfloor\frac{n}{dt}\right\rfloor\\ &=\sum_{T|k}\left\lfloor\frac{n}{T}\right\rfloor\sum_{d|T}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor) \end{aligned}\]

所以就有了$f_i=\frac{\sum\limits_{T|k}\left\lfloor\frac{n}{T}\right\rfloor\sum\limits_{d|T}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor)}{n}+1$

再考虑把整个柿子当成一个以$f_i$为未知量的方程解。

\[\begin{aligned} f_i&=\frac{\sum\limits_{T|k}\left\lfloor\frac{n}{T}\right\rfloor\sum\limits_{d|T}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor)}{n}+1\\ f_i&=\frac{\sum\limits_{T|k}\left\lfloor\frac{n}{T}\right\rfloor\sum\limits_{d|T,d<i}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor)}{n}+1+\frac{\left\lfloor\frac{n}{i}\right\rfloor\times f_i}{n}\\ f_i&=\frac{\sum\limits_{T|k}\left\lfloor\frac{n}{T}\right\rfloor\sum\limits_{d|T,d<i}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor)+n}{n-\left\lfloor\frac{n}{i}\right\rfloor} \end{aligned}\]

然后设$dp_i=\sum\limits_{d|T,d<i}f_d\mu(\left\lfloor\frac{T}{d}\right\rfloor)$。

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……

posted @ 2024-10-15 21:27 CuFeO4 阅读(38) 评论(0) 编辑收藏举报

刷新页面返回顶部

Cu

vjudge数学专题1

F New Year and Arbitrary Arrangement

E - Steps to One

公告