2020 HDU 杭电多校第六场

2020 HDU 杭电多校第场六场

1. Little Rabbit’s Equation
2. Divisibility
3. A Very Easy Graph Problem
4. Road To The 3rd Building
5. Expectation
6. Fragrant numbers

Little Rabbit’s Equation

题目大意： 给一串数字符号表达式,形如"A1+30=B5" 判断这个最少是几进制的，

由于数据很小，直接模拟，注意每个数字必须严格小于进制数

#include 
#define rep(i,n) for (int i = 0 ; i < (n) ; ++i)
#define repn(i,n) for (int i = 1; i <= (n) ; ++i)
#define rep2(i,j,n) for (int i = (j) ; i < (n) ; ++i)
#define repn2(i,j,n) for (int i = (j); i <= (n) ; ++i)
#define debug(x) printf("%s = %d\n",#x,x)
#define pb push_back
#define mk make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;

ll qpow(ll a , ll exp ,ll mod = mod){ ll res = 1; while(exp){ if(exp & 1) res = res * a % mod; a = a * a % mod; exp >>= 1;} return res; }
ll inv(ll num , ll mod = mod){ return qpow(num,mod-2);}

#define int ll
struct Moder{
    template<typename T> static T add(const T & a ,const T& b){ return ((a+b) % mod + mod) % mod;}
    template<typename T> static T mul(const T & a ,const T& b){ return ((a-b) % mod + mod) % mod;}
    template<typename T> static T get(const T & a){ return (a % mod + mod) % mod;}
};
struct FIO{ // char*s, int
    int c,f,e;  operator int() { return ~e; }
    FIO& operator >> (char* s){e=scanf("%s",s);return *this;}
    template<class T>FIO& operator >> (T& x){
        for(f=0;~c&&(c<48||c>57);c=getchar()){if(c=='-')f=1;}
        for(x=0,e=c;c>47 && c<58;c=getchar()){x=x*10+(c^48);}
        if(f) {x=-x;}   return *this;}
};
string s[3];
ll num[3];

int op;
int isOper(char c){
    if(c == '+') return 1;
    if(c == '-') return 2;
    if(c == '*') return 3;
    if(c == '/') return 4;
    return 0;
}

void split(string str){
    int p = 0 , i = 0;
    int siz = str.size();
     
    rep(i,3) s[i] = "";
    
    while(! (op = isOper(str[i])) ) {
        s[p] = s[p] + str[i];
        i++;
    }
    i++; p++;
    while(str[i] != '='){
        s[p] = s[p] + str[i];
        i++;
    } i++; p++;
    while(i < siz){
        s[p] = s[p] + str[i];
        i++;
    } 

}

int getnum(char c){
    if(isdigit(c)) return c-'0';
    else return c - 'A' + 10;
}

bool getNum(int id , ll bs){
    int len = s[id].size();
    ll & tmp = num[id];
    tmp = 0;
    
    rep(i,len){
        int digit = getnum(s[id][i]);
        if(digit >= bs) return 0;
        tmp = tmp * bs + digit;
    }    
    return true;
}

bool test(ll bs){
    rep(i,3) if(!getNum(i,bs)) return false;
    if(op == 1){
        return num[0] + num[1] == num[2];
    } else if (op == 2){
        return num[0] - num[1] == num[2];
    } else if (op == 3){
        return num[0] * num[1] == num[2];
    } else {
        return (num[0] / num[1] == num[2]) && (num[0] % num[1] == 0); 
    }    
}
string ss;
signed main()
{
    while(cin >> ss){
        split(ss);
        int res = -1;
        for (int i = 2 ; i <= 16 ; ++i){
            if(test(i)){
                res = i;
                break;
            }
        }
        cout << res << endl;        
    }

    return 0;
}

Divisbility

题目大意：
比如能被3整除的数在10进制下有个特点：和能被3整除
给定一个进制和数字，判断是否也有这样类似的性质。

$S=a_0+10a_1+100a_2+.........10^n{a}_n$

10进制下为什么会有这个性质呢？
我们让S mod 3 得到如下

$S\mathrm{m}\mathrm{o}\mathrm{d}3=a_0+a_1+a_2+........a_n$

可以看出之所以会有这个性质，是因为10的任意次方，对3取余的结果都是1，正因如此，整个式子能被3整除即代表了各个位置和能被3整除

也就是说 $(Base-1)\mathrm{m}\mathrm{o}\mathrm{d}X=0$ 成立

#include 
#define rep(i,n) for (int i = 0 ; i < (n) ; ++i)
#define repn(i,n) for (int i = 1; i <= (n) ; ++i)
#define rep2(i,j,n) for (int i = (j) ; i < (n) ; ++i)
#define repn2(i,j,n) for (int i = (j); i <= (n) ; ++i)
#define debug(x) printf("%s = %d\n",#x,x)
#define pb push_back
#define mk make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;

ll qpow(ll a , ll exp ,ll mod = mod){ ll res = 1; while(exp){ if(exp & 1) res = res * a % mod; a = a * a % mod; exp >>= 1;} return res; }
ll inv(ll num , ll mod = mod){ return qpow(num,mod-2);}

struct Moder{
    template<typename T> static T add(const T & a ,const T& b){ return ((a+b) % mod + mod) % mod;}
    template<typename T> static T mul(const T & a ,const T& b){ return ((a-b) % mod + mod) % mod;}
    template<typename T> static T get(const T & a){ return (a % mod + mod) % mod;}
};
struct FIO{ // char*s, int
    int c,f,e;  operator int() { return ~e; }
    FIO& operator >> (char* s){e=scanf("%s",s);return *this;}
    template<class T>FIO& operator >> (T& x){
        for(f=0;~c&&(c<48||c>57);c=getchar()){if(c=='-')f=1;}
        for(x=0,e=c;c>47 && c<58;c=getchar()){x=x*10+(c^48);}
        if(f) {x=-x;}   return *this;}
};

int Case;
ll b,x;
int main()
{
    scanf("%d",&Case);
    while(Case--){
        scanf("%lld%lld",&b,&x);
        if((b-1) % x == 0)puts("T");
        else puts("F");    
    }    

    return 0;
}

A Very Easy Graph Problem

题目大意：
给出一张无向图，给出所有顶点的权值V $\{0,1\}$
求所有任意$V_a\oplus V_b=1$ 的两点，最短路径和（值计算一次）
$V\le 1e5,E\le 2e5$

如果题只读到这是根本做不下去的，我一开始忽略了很重要的一句话

第i条边的长度为：$2^i$

也就是说：即便你用尽$1......(i-1)$所有的边，都不可能超过第$i$条长度

于是考虑用并查集维护图的连通性，如果图已经联通，就不考虑后面加入的边（因为肯定不会用这条边），这样一来，我们可以保证最后我们的图只有n-1边，从无向图变成了一颗数。

有道“点分治”模板题是求书上所有点对距离，可以稍作修改
每次分治，分别处理每颗子树，记录序号为0和1的总长度，和记号为0，1的所有点的个数

//sc0: 记录所有子树，权值为0节点总和，在一颗子树搜索完后更新，因为
//点分每一步是考虑跨过根节点的，所以用临时变量存同一颗子树的信息，结束后更新
void dfs(int x,  ll cost , int f=-1){
    cost %= mod;
    if(arr[x] == 0){
        c0 += cost;		//c0表示权值为0子树结点长度总和
        o0++;			//o0用于计数
        if(cnt1) ans += sc1 + cnt1*cost;
    }
    else {
        c1 += cost;
        o1++;
        if(cnt0) ans += sc0 + cnt0*cost; 
    }
    ans %= mod;
        
    for (Edge& e : G[x]){
        int v = e.v;
        if(vis[v] || v == f) continue;
        dfs(v,e.cost+cost,x);
    }
} 

详细处理代码如下

#include 
#define rep(i,n) for (int i = 0 ; i < (n) ; ++i)
#define repn(i,n) for (int i = 1; i <= (n) ; ++i)
#define rep2(i,j,n) for (int i = (j) ; i < (n) ; ++i)
#define repn2(i,j,n) for (int i = (j); i <= (n) ; ++i)
#define debug(x) printf("%s = %d\n",#x,x)
#define pb push_back
#define mk make_pair
#define fi first
#define eb emplace_back
#define se second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;

ll qpow(ll a , ll exp ,ll mod = mod){ ll res = 1; while(exp){ if(exp & 1) res = res * a % mod; a = a * a % mod; exp >>= 1;} return res; }
ll inv(ll num , ll mod = mod){ return qpow(num,mod-2);}
#define int ll
struct Moder{
    template<typename T> static T add(const T & a ,const T& b){ return ((a+b) % mod + mod) % mod;}
    template<typename T> static T mul(const T & a ,const T& b){ return ((a-b) % mod + mod) % mod;}
    template<typename T> static T get(const T & a){ return (a % mod + mod) % mod;}
};
struct FIO{ // char*s, int
    int c,f,e;  operator int() { return ~e; }
    FIO& operator >> (char* s){e=scanf("%s",s);return *this;}
    template<class T>FIO& operator >> (T& x){
        for(f=0;~c&&(c<48||c>57);c=getchar()){if(c=='-')f=1;}
        for(x=0,e=c;c>47 && c<58;c=getchar()){x=x*10+(c^48);}
        if(f) {x=-x;}   return *this;}
};
const int maxn = 4e5 + 100;
int Case;
int Size;
int n,m,x,y;
int arr[maxn];
int MX;
struct Edge{
    int v;
    ll cost;
};
vector<Edge> G[maxn];

int siz[maxn], mxson[maxn] , root ,ans;
bool vis[maxn];
int fa[maxn];

void init(const int n){
    rep(i,n+2) fa[i] = i;
    rep(i,n+2) siz[i] = mxson[i] = vis[i] = 0;
    repn(i,n) G[i].clear();
    root = ans = 0;
}
int Find(int x){  return fa[x] == x ? x : fa[x] = Find(fa[x]);}
 
void getroot(int u,int f = -1){
    siz[u] = 1; mxson[u] = 0;
    for (Edge & e : G[u]){
        int v = e.v;
        if(f == v || vis[v]) continue;
        getroot(v,u);
        siz[u] += siz[v];
        mxson[u] = max(siz[v],mxson[u]);
    }
    mxson[u] = max(mxson[u],Size-siz[u]);
    if(mxson[u] < MX){
        root = u;
        MX = mxson[u];
    } 
}
ll cnt0,cnt1;
ll sc0, sc1;
ll c0,c1;
ll o0,o1;


void dfs(int x,  ll cost , int f=-1){
    cost %= mod;
    if(arr[x] == 0){
        c0 += cost;
        o0++;
        if(cnt1) ans += sc1 + cnt1*cost;
    }
    else {
        c1 += cost;
        o1++;
        if(cnt0) ans += sc0 + cnt0*cost; 
    }
    ans %= mod;
        
    for (Edge& e : G[x]){
        int v = e.v;
        if(vis[v] || v == f) continue;
        dfs(v,e.cost+cost,x);
    }
} 

void Divide(int rt , int ssize){

    sc0 = 0 , sc1 = 0;
    cnt0 = cnt1 = 0;

    vis[rt] = 1;
    for (Edge& e : G[rt]){
        int v = e.v;
        if(vis[v]) continue;
        c0 = c1 = o0 = o1 = 0;
        dfs(v,e.cost,rt);
        
        sc0  += c0;  sc1 += c1;
        sc0  %= mod; sc1 %= mod;
        
        cnt0 += o0; cnt1 += o1;
    }
    
    if(arr[rt] == 0) ans += sc1;
    else ans += sc0;
    ans %= mod; 
    
    for (Edge& e : G[rt]){
        int v = e.v;
        if(vis[v]) continue;
        
        Size = siz[v];
        MX = inf; getroot(v,0);
        Divide(root,Size);
    }
    
}

signed main()
{
    scanf("%lld",&Case);

    while(Case--){
        scanf("%lld%lld",&n,&m);    
        init(n);
        ll len = 1;
        ans = 0;
    
        repn(i,n) scanf("%lld",arr+i);
    
        Size = n;
        repn(i,m){
            scanf("%lld%lld",&x,&y);
            len <<= 1;
            len %= mod;
            int fx = Find(x), fy = Find(y);
            if(fx == fy) continue;
            else fa[fx] = fy;
            
            G[x].eb(Edge{y,len});
            G[y].eb(Edge{x,len}); 
            
        }
        root = 0;MX = inf;  getroot(1,-1);
        Divide(root,n);
        cout << ans << endl;
        
    }    

    return 0;
}

Road To The 3rd Building

题目大意：任取 $i,j,1\le i\le j\le n$，求${\sum }_{k=i}^j{a}_k/(j-i+1)$的数学期望

做题时找灵感想到了滑动窗口，比如一个长度为7的数组，
枚举窗口长度和每个数字出现的额次数

$len=1$
$1111111$

$len=2$
$1222221$

$len=3$
$1233321$

$len=4$
$1234321$

$len=5$
$1233321$

$len=6$
$1222221$

$len=7$
$1111111$

每次起加点往左一格，结束点往右一格，超过一半后反过来，这就很好做了。

const int maxn = 1e6 + 100; 
int Case , n;
ll sum[maxn];

signed main()
{
    scanf("%lld",&Case);
    while(Case--){
        scanf("%lld",&n);
        repn(i,n) scanf("%lld",sum+i);
        repn(i,n) sum[i] += sum[i-1];
        
        ll d = n*(n+1) / 2; d %= mod;
        d =  inv(d) % mod;
        
        ll res = 0 , add = 0,  div_inv = 0;
        
        int bd = (n+1)  / 2;
        
        for (int i = 1 ; i <= bd ; ++i){
            add += sum[n-i+1] - sum[i-1];
            add %= mod;
            div_inv = inv(i);
            res += add * div_inv % mod;
            res %= mod;
        }
        
        add = 0;
        for (int i = 1 ; i <= n - bd ; ++i){
            add += sum[n-i+1] - sum[i-1];
            add %= mod;
            div_inv = inv(n-i+1);
            res += add * div_inv % mod;
            res %= mod;
        }
        
        res = res * d % mod;
        cout << res << endl;
    }

    return 0;
}

Expectation

题目大意: 定义生成树的权值为边权的“与和“。
给点边和点，随机生成一个生成树，求生成树权值的期望。

首先我们要确定一件事情：
“与和“的期望 等于 每一位”与“的期望之和

因此考虑贪心方法：建图，每次只考虑第$i$位是1的边，进行计算当前生成树数量，这样我们就可以得到：

$E(G)={\sum }_{i=1}^n{2}^i\ast Count(\{edge|edge.c\&2^i==1\})/Count(\{edge\})$

问题就在于如何求生成树个数了

有个$Matrix-tree$定理，就是关于这个的，这个我也是只知道，只会拿来用，详情可以看看别人的博客

#include 
#define rep(i,n) for (int i = 0 ; i < (n) ; ++i)
#define repn(i,n) for (int i = 1; i <= (n) ; ++i)
#define rep2(i,j,n) for (int i = (j) ; i < (n) ; ++i)
#define repn2(i,j,n) for (int i = (j); i <= (n) ; ++i)
#define debug(x) printf("%s = %d\n",#x,x)
#define pb push_back
#define mk make_pair
#define fi first
#define eb emplace_back
#define se second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
#define int ll
ll qpow(ll a , ll exp ,ll mod = mod){ ll res = 1; while(exp){ if(exp & 1) res = res * a % mod; a = a * a % mod; exp >>= 1;} return res; }
ll inv(ll num , ll mod = mod){ return qpow(num,mod-2);}

struct Moder{
    template<typename T> static T add(const T & a ,const T& b){ return ((a+b) % mod + mod) % mod;}
    template<typename T> static T mul(const T & a ,const T& b){ return ((a-b) % mod + mod) % mod;}
    template<typename T> static T get(const T & a){ return (a % mod + mod) % mod;}
};
struct FIO{ // char*s, int
    int c,f,e;  operator int() { return ~e; }
    FIO& operator >> (char* s){e=scanf("%s",s);return *this;}
    template<class T>FIO& operator >> (T& x){
        for(f=0;~c&&(c<48||c>57);c=getchar()){if(c=='-')f=1;}
        for(x=0,e=c;c>47 && c<58;c=getchar()){x=x*10+(c^48);}
        if(f) {x=-x;}   return *this;}
};

const int maxn = 2e4+ 100;; 
#define LL ll
int Case;
//图论部分
struct Edge{
    int u,v;
    ll cost;
}; 

const int N = 111;
LL K[N][N];
LL gauss(int n){
    LL res=1;
    for(int i=1;i<=n-1;i++){
        for(int j=i+1;j<=n-1;j++){
            while(K[j][i]){
                int t=K[i][i]/K[j][i];
                for(int k=i;k<=n-1;k++)
                    K[i][k]=(K[i][k]-t*K[j][k]+mod)%mod;
                swap(K[i],K[j]);
                res=-res;
            }
        }
        res=(res*K[i][i])%mod;
    }
    return (res + mod) % mod;
}


int n,m;

vector<Edge> edge;

void build(int x){
    memset(K,0,sizeof(K));
    for(Edge&e : edge){
        int u = e.u , v = e.v;
        ll c = e.cost;
        if((c>>x) & 1){
            K[u][u]++; K[v][v]++;
            K[u][v]--; K[v][u]--;
        }
    }
}

signed main()
{
    scanf("%lld",&Case);
    while(Case--){
        edge.clear();
        memset(K,0,sizeof(K));
        
        ll total = 0; 
        int x,y,z;
        
        scanf("%lld%lld",&n,&m);
        repn(i,m){
            scanf("%lld%lld%lld",&x,&y,&z);
            edge.eb(Edge{x,y,z});
            K[x][x]++; K[y][y]++;
            K[x][y]--; K[y][x]--;     
        }
        total = gauss(n);
        ll inv_t = inv(total); 
        ll res = 0;
        for (int i = 30 ; i >= 0 ; --i){
            ll base = 1LL << i;
            build(i);
            
            ll cnt = gauss(n);
            
            ll tmp = base * cnt % mod * inv_t % mod;
            res = (res + tmp) % mod;
        }
        cout << res << endl; 
    }
    
    return 0;
}

Fragrant numbers

题目大意：（太 臭 了）
一个无限序列114514919114514…
可以用$+，\ast$和括号来连接,给出t个询问，
问想要获得数$N\le 5000$，至少需要多长的序列（序列从头开始）

做法:…暴力跑了dp10位后…发现3，7不行，其余都行。。。。直接交了

（做法太臭了)

#include 
#define rep(i,n) for (int i = 0 ; i < (n) ; ++i)
#define repn(i,n) for (int i = 1; i <= (n) ; ++i)
#define rep2(i,j,n) for (int i = (j) ; i < (n) ; ++i)
#define repn2(i,j,n) for (int i = (j); i <= (n) ; ++i)
#define debug(x) printf("%s = %d\n",#x,x)
#define pb push_back
#define mk make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;

ll qpow(ll a , ll exp ,ll mod = mod){ ll res = 1; while(exp){ if(exp & 1) res = res * a % mod; a = a * a % mod; exp >>= 1;} return res; }
ll inv(ll num , ll mod = mod){ return qpow(num,mod-2);}

struct Moder{
    template<typename T> static T add(const T & a ,const T& b){ return ((a+b) % mod + mod) % mod;}
    template<typename T> static T mul(const T & a ,const T& b){ return ((a-b) % mod + mod) % mod;}
    template<typename T> static T get(const T & a){ return (a % mod + mod) % mod;}
};
struct FIO{ // char*s, int
    int c,f,e;  operator int() { return ~e; }
    FIO& operator >> (char* s){e=scanf("%s",s);return *this;}
    template<class T>FIO& operator >> (T& x){
        for(f=0;~c&&(c<48||c>57);c=getchar()){if(c=='-')f=1;}
        for(x=0,e=c;c>47 && c<58;c=getchar()){x=x*10+(c^48);}
        if(f) {x=-x;}   return *this;}
};
const int maxn =155;
const int bb = 5000;
set<int> F[maxn][maxn];
bool vis[maxn][maxn];
int arr[10] = {1,1,4,5,1,4,1,9,1,9};

int con(int a , int b){
    int t = 10;
    int tb = 0;
    while(tb){
        t *= 10;
        tb /= 10;
    }
    return a * t + b;
}

set<int> & dp (int l , int r){
    if(vis[l][r]) return F[l][r];
    vis[l][r] = 1;
    set<int> & tmp = F[l][r];
        
    if(l == r){
        int digit = arr[l%10];    
        tmp.insert(digit);
        return tmp;
    }
    int p = l;
    int sum = 0;
    
    while(p <= r){
        sum = sum * 10 + arr[p % 10];
        if(sum > 5000) break; 
        p++;
    }
    if(sum <= 5000) tmp.insert(sum);


    for(int i = l ; i < r ; ++i){
        
        set<int> &left = dp(l,i);
        set<int> &right = dp(i+1,r);
        
        if(l == 0 && r == 5 &&  i == 2){
            
            int p = 1;
            p++;
        }
        for (int lv : left) for (int rv : right){
            if(lv * rv <= 5000){
                tmp.insert(lv * rv);
            } 
            if(lv + rv <= 5000){
                tmp.insert(lv + rv);
            }
        }
    } 
    return tmp;
} 
map<int,int> mps;
int main()
{
    int Case; 
    FIO read;
    dp(0,10);
    int p = 0;
    
    
    for (int i = 0 ; i <= 10 ; ++i)
    
    for (int v : F[0][i]) {
        if(mps[v] == 0) {
            //cout << v << ' ';
            mps[v] = i + 1; 
        }
    }
    mps[0] = 0;
    

    scanf("%d",&Case);
    while(Case--){
        int p ;
        cin >> p;
        if(mps.find(p) == mps.end()) puts("-1");
        else cout << mps[p] << endl;
    }
    

    return 0;
}