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csp模拟18[最长反链, 2A + X, No Rest for the Wicked , 暴力题]

时间:2022-10-13 16:33:37浏览次数:45  
标签:rt No int 18 LL Rest read Mod define

csp模拟18[最长反链, 2A + X, No Rest for the Wicked , 暴力题]

T1 最长反链

明确两件事 :

  • 位数相同的数不会冲突。
  • 一个高位数,会与它各个位上所能组成的数冲突。
  • 所以这个东西是有很明显的上下界的,我们考虑一个性质,对于高位\(\geq 2\) 的数字来说,它的下界就是它的量级, \([25, 156]\)这种来说,下界就可以是\(57\)否。
  • 考虑一种解法则是有两种来源(eg \(231\)下界是\(100\), \([25, 156]\)这种来说,下界就可以是\(57\))。
  • 考虑一种解法
    • 对于r,我们可以拆出两个只比它小一位的数,即\(r / 10\) 和 \(r\) 去掉它的最高位组成的数
    • 这两个数显然是不能选的, 且比这两个数中较大的数小的数显然可以在与\(r\)位数相同的数中找到, 所以剩下的只有比\(\ max\)(\(l\), \(max\)(\(r\)中选出的数))更大, 比\(r\)更小的数讨论。剩下分讨就行。
here


#include <bits/stdc++.h>
#define LL long long
#define Re register int
#define LD double
#define mes(x, y) memset(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; ++ x)
#define fp(x, y, z)for(Re x = y; x >= z; -- x)
#define delfr(x, y, z)for(Re x = y; x < z; ++ x)
#define delfp(x, y, z)for(Re x = y; x > z; -- x)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define mk(x, y) make_pair(x, y)
#define pb(x) push_back(x)
#define sec second
#define fst first
using namespace std;
namespace kiritokazuto{
    // auto read = [](){
    //     LL x = 0;
    //     int f = 1;
    //     char c;
    //     while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
    //     do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
    //     return x * f;
    // };
    inline char getc(){
        static char buf[1<<18],*p1,*p2;
        if(p1==p2){p1=buf,p2=buf+fread(buf,1,1<<18,stdin);if(p1==p2)return EOF;}
        return *p1++;
    }
    inline LL read(){
        LL x=0;char ch=getc();
        while(!isdigit(ch))ch=getc();
        while(isdigit(ch))x=(x<<1)+(x<<3)+(ch^48),ch=getc();
        return x;
    }
    template <typename T> fuc(void, write) (T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10);putchar(x % 10 | '0');
    }

}

using namespace kiritokazuto;
const int maxn = 3e5 + 10, Inf = 2147483647, Mod = 998244353;
int t;
#define int long long
LL x, y;
fuc(int, getlj)(LL x) {//得到量级
    if(x / 1000000000)return 9;
    if(x / 100000000)return 8;
    if(x / 10000000)return 7;
    if(x / 1000000)return 6;
    if(x / 100000)return 5;
    if(x / 10000)return 4;
    if(x / 1000)return 3;
    if(x / 100)return 2;
    if(x / 10)return 1;
    if(x || !x)return 0;
}

fuc(int, get)(LL x) {
    if(x < 10)return x;
    return get(x / 10) * 10;
}
LL big;
LL ljx, ljy;
LL lj[12] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};
signed main(){
   t = read();
   while(t --) {
       x = read(), y = read();
       if(getlj(x) == getlj(y)) {
           write(y - x + 1);
           ki;
           continue;
       }
       ljx = getlj(x) + 1, ljy = getlj(y) + 1;
       if(y >= lj[ljy - 1] * 2 - 1) {
           write(y - lj[ljy - 1] + 1);
           ki;
           continue;
       }
       x = max(x, lj[ljy - 2] );
       int l = max(y / 10, y % lj[ljy - 1]) +1;
       write(y - max(l, x) + 1);
       ki;
       
   }
   return 0;
}
 

    

/*


3
3 8
3 18
1 1000


6
10
900


*/

T2 2A + X

  • 我们只有两种操作,一种是乘二,一种是加一,在不同时刻的加一相当于是\(+1, +2, +4, +8 ...\)不等的操作,我们可以考虑一个事,因为要极差最小,一个很显然的思路就是通过这些操作不断使最小值逼近最大值。
  • 我们可以考虑使一个数一直乘二,不断去逼近最大值,并且记录其乘二的次数,通过乘二的次数我们可以推出它能加哪些东西,再次逼近。
  • 同时我们还需要考虑一个事情,就是最大值的改变,所以我们维护一个\(nxt\)表示当前的数加上一个它所能加的最小的值超过\(Max\)之后的值,这样最后统计答案时就可以先让它做最小值(排序后),让他变大,让\(i + 1\)做最小值,然后\(O(n)\)推出答案了,目前我在题库里跑的最快
  • 当然还有一些东西,如果一个数已经和最大值一样了,或者乘二超过了最大值,我们把它的\(nxt\)直接改成\(Max\)或者是\(a_{i} \times 2\)就行。
  • 还需要特判一些\(0\)的情况,就是差值小于\(x\),则显然可以跟上。
here


#include <bits/stdc++.h>
#define LL long long
#define Re register int
#define LD double
#define mes(x, y) memset(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; ++ x)
#define fp(x, y, z)for(Re x = y; x >= z; -- x)
#define delfr(x, y, z)for(Re x = y; x < z; ++ x)
#define delfp(x, y, z)for(Re x = y; x > z; -- x)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define mk(x, y) make_pair(x, y)
#define pb(x) push_back(x)
#define sec second
#define fst first
 
using namespace std;
namespace kiritokazuto{
    // auto read = [](){
    //     LL x = 0;
    //     int f = 1;
    //     char c;
    //     while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
    //     do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
    //     return x * f;
    // };
    inline char getc(){
        static char buf[1<<18],*p1,*p2;
        if(p1==p2){p1=buf,p2=buf+fread(buf,1,1<<18,stdin);if(p1==p2)return EOF;}
        return *p1++;
    }
    inline LL read(){
        LL x=0;char ch=getc();
        while(!isdigit(ch))ch=getc();
        while(isdigit(ch))x=(x<<1)+(x<<3)+(ch^48),ch=getc();
        return x;
    }
    template <typename T> fuc(void, write) (T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10);putchar(x % 10 | '0');
    }
 
}   
 
using namespace kiritokazuto;
const int maxn = 5e5 + 10, Inf = 2147483647, Mod = 1e9 + 7;

int n;
int x;
int a[maxn];
int cnt;
struct Node {
    int val , nxt, id;
    friend bool operator < (Node A, Node B) {
        return A.val < B.val;
    }
}b[maxn];
int num;
int Max = -1;
int Mincard = Inf;
int Min;
signed main(){
   n = read();
   x = read();
   fr(i, 1, n)a[i] = read(), Max = max(Max, a[i]);
   fr(i ,1 ,n) {
       b[i].nxt = cnt = -1;//2 ^ 0 = 1所以从-1存,这里nxt临时存成了几个二
       while((a[i] << 1) <= Max) {
           cnt++;
           a[i] <<= 1;
       }
       b[i].val = a[i];
       b[i].id = i;
    //    printf("a[%d] = %d cnt = %d\n", i, a[i], cnt);
       for(Re j = cnt; ~j; j --) {//枚举出牌
            if(b[i].val == Max)continue;
            int tmp = Max - b[i].val;
            num = min(tmp / (1 << j), x);
            b[i].val += num * (1 << j);//因为从-1存,所以这里不减一
            if(num != x) b[i].nxt = j;//最小卡
       }
   }
//    fr(i, 1, n) {
//        printf("b[%d].val = %d b[%d].nxt = %d a[%d] = %d\n", i, b[i].val, i, b[i].nxt, i, a[i]);
//    }
   sort(b + 1, b + n + 1);
   if(b[n].val - b[1].val < x) {
       puts("0");
       return 0;
   }
   Min = Max - b[1].val;
//    printf("%d \n", Min);
   delfr(i, 1, n) {
       if(b[i].val == b[n].val) {b[i].nxt = b[i].val;continue;}
       if(~b[i].nxt)b[i].nxt = min(b[i].val + (1 << b[i].nxt), a[b[i].id] << 1);
       else {
        //    if(b[i].val == 12) {
        //        printf("%d\n", a[b[i].id]);
        //    }
           b[i].nxt = (a[b[i].id] << 1);
       }
   }
//    delfr(i, 1, n) {
//        printf("b[%d].val = %d b[%d].nxt = %d\n", i, b[i].val, i, b[i].nxt);
//    }
   delfr(i, 1, n) {
       Max = max(Max, b[i].nxt);
       Min = min(Min, Max - b[i + 1].val);
   }
   write(Min);

   return 0;
}

T3 No Rest for the Wicked

  • 这个东西就跟着官方题解走呗,我们设\(f(x, v)\)表示从\(v\)出发,最大疫情值是\(x\)的\(ans\),显然我们可以由疫情值小的走到大的,那么小的就可以继承大的的答案,直接不用搜,停止就行。当然,如果仅仅是这样,显然我们可以对\(c_{i}\)排个序,然后\(bfs\)"记忆化”一下,然后就可以获得\(67pts\)的好成绩,不过这样仍然是\(O(n ^ 2)\)的。
  • 考虑优化,对于两个点(\(u, v\))
    • 如果\(\max(c_{u}, c_{v}) \le x \le \min(t_{v}, t_{u})\)那么显然\(x\)在这两个点之间走的时候是不会更改的,所以他们相当于是一个联通块,答案可以共享。
    • 如果\(c_{u} \le x \le \min(t_{u}, c_{v} - 1)\),那么显然我们可以走到\(u\),也可以从\(u\)走到\(v\)。
      • 首先考虑\(min\)取到\(c_{v} - 1\), 走到\(v\)之后我们的\(x\)会变大,所以此时建造双向边是无用的,因为我们在针对\(x\)处理,对于这些边是用来更新答案的,从小的疫情值搜到大的疫情值,我们可以直接继承大的的答案,所以显然没有从大的再到小的的必要,建造单向边,双向回来的话是针对\(x\)等于\(c_{v}\)之后,也就是到达\(v\)再考虑的,如果双向边可以建我是会建的(相当于线段树分治板子里时间还未到c_{v}仅仅只到了\(x\))
      • 此外如果\(min\)取到\(t_{u}\)显然是回不去了,单向边。
    • 考虑上文的边对于不同的\(x\)是可能存在或者消失的,这就符合线段树分治这种十分友好的离线算法了
    • 我们可以对\(c\)进行分治,递归到叶子节点时统计答案,维护一个\(val\)和\(ans\)取一个\(max\)即可。
here




#include <bits/stdc++.h>

#define LL long long
#define Re register int
#define LD double
#define mes(x, y) memset(x, y, sizeof(x))
#define cpt(x, y) memcpy(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; ++ x)
#define fp(x, y, z)for(Re x = y; x >= z; -- x)
#define delfr(x, y, z)for(Re x = y; x < z; ++ x)
#define delfp(x, y, z)for(Re x = y; x > z; -- x)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define mk(x, y) make_pair(x, y)
#define pb(x) push_back(x)
#define re(x) return x
#define sec second
#define fst first
using namespace std;
namespace kiritokazuto{
    auto read = [](){
        LL x = 0;
        int f = 1;
        char c;
        while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
        do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
        return x * f;
    };
    template <typename T> fuc(void, write) (T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10);putchar(x % 10 | '0');
    }

}

using namespace kiritokazuto;
const int maxn = 4e5 + 10, Inf = 0x3f3f3f3f, Mod = 5000007;
#define int long long
struct City {
    int c, s, t;
    friend bool operator < (City A, City B) {
        return A.c > B.c;
    }
}ct[maxn];
int n, m, k;
int lsh[maxn], cnt;
struct EDGE{int from, to;}wmx[maxn];
struct STK{int x, y, valf, tim;bool type;}stk[maxn];
//x是爹y是儿子val是爹的val
int top = 0;
int fa[maxn];
vector <int> qr[maxn << 1];
int heigh[maxn];
int ans[maxn], val[maxn];
struct Set_Union {
    fuc(void, init)(){fr(i, 1, n)fa[i] = i, heigh[i] = 1, val[i] = ct[i].s, ans[i] = ct[i].s;}
    fuc(int , find)(int x) {while(x != fa[x])x = fa[x]; return x;}//不进行路径压缩
    fuc(void, merge)(int x, int y) {
        x = find(x), y = find(y);
        if(x == y)return;
        if(heigh[x] < heigh[y])swap(x, y);//按秩合并
        fa[y] = x;
        heigh[x] += heigh[y];//小的进大的
        val[x] = max(val[x], val[y]);
    }
    fuc(void, add)(int tim, int pos, int type) {
        if(type) {
            int from = wmx[pos].from, to = wmx[pos].to;
            from = find(from), to = find(to);
            if(val[from] < ans[to]) {
                stk[++top].tim  = tim;
                stk[top].type = type;
                stk[top].x = stk[top].y = from;
                stk[top].valf = val[from];
                val[from] = ans[to];
            }
        }else {
            int from = wmx[pos].from, to = wmx[pos].to;
            from = find(from), to = find(to);
            if(from == to)return;
            if(heigh[from] < heigh[to])swap(from, to);
            stk[++top].x = from;
            stk[top].y = to;
            stk[top].valf = val[from];
            stk[top].type = type;
            stk[top].tim = tim;
            fa[to] = from;
            val[from] = max(val[from], val[to]);
            heigh[from] += heigh[to];
        }
    }
    fuc(void, del)(int tim) {
        while(top && stk[top].tim == tim) {
            if(stk[top].type) {
                val[stk[top].y] = stk[top].valf;
            }else {
                int x = stk[top].x, y = stk[top].y;
                heigh[x] -= heigh[y];
                fa[y] = y;
                val[x] = stk[top].valf;
            }
            top--;
        }
    }
}F;
struct Seg_Tree {
    #define lsp(rt) (rt << 1)
    #define rsp(rt) (rt << 1 | 1)
    #define mid ((l + r) >> 1)
    struct Tree {
        vector <int> sig, Two;
    }tr[5000001];
    fuc(void, insert)(int rt, int l, int r, int L, int R, int pos, int type) {
        if(L <= l && r <= R) {
            if(type)tr[rt].sig.pb(pos);
            else tr[rt].Two.pb(pos);
            return;
        }
        if(L <= mid)insert(lsp(rt), l, mid, L, R, pos, type);
        if(R > mid)insert(rsp(rt), mid + 1, r, L, R, pos, type);
    }
    fuc(void, query)(int rt, int l, int r) {
        for(auto it : tr[rt].sig)F.add(rt, it, 1);
        for(auto it : tr[rt].Two)F.add(rt, it, 0);
        if(l == r) {//叶子
            for(auto it : qr[l])ans[it] = max(ans[it], val[F.find(it)]);
            return F.del(rt), void();
        }
        query(rsp(rt), mid + 1, r);
        query(lsp(rt), l, mid);
        F.del(rt);

    }
}T;

signed main(){
    n = read();
    m = read();
    fr(i, 1, n) ct[i].c = read() ,ct[i].t = read() ,ct[i].s = read(), lsh[++cnt] = ct[i].c, lsh[++cnt] = ct[i].t;
    fr(i, 1, m)wmx[i].from = read(), wmx[i].to = read();
    sort(lsh + 1, lsh + cnt + 1);
    cnt = unique(lsh + 1, lsh + cnt + 1) - lsh - 1;
    fr(i, 1, n) {
        ct[i].c = lower_bound(lsh + 1, lsh + cnt + 1, ct[i].c) - lsh;
        ct[i].t = lower_bound(lsh + 1, lsh + cnt + 1, ct[i].t) - lsh;
        qr[ct[i].c].pb(i);
    }
 fr(i, 1, m) {
        int &from = wmx[i].from, &to = wmx[i].to;
        if(ct[from].c > ct[to].c)swap(from, to);
        T.insert(1, 1, cnt, ct[to].c, min(ct[from].t, ct[to].t), i, 0);
        if(ct[from].c <= min(ct[to].c - 1, ct[from].t))T.insert(1, 1, cnt, ct[from].c, min(ct[to].c - 1, ct[from].t), i, 1);
    }
    F.init();
    T.query(1, 1, cnt);
    fr(i, 1, n)write(ans[i]), fk;
 return 0;
}

T4 暴力题

  • 对于题目的柿子我们可以去掉下取整,变为

\[\sum \limits_{i = 1}^{n} \frac{b_{i} \times i ^ {k} - (b^{i} \times i ^ {k})\% w}{w} \]

  • 然后就可以权值线段树维护这两个东西了,对于除以\(w\)用乘逆元实现就行。
  • 我们设\(f(x)\)表示\(\sum\limits_{i = 1} ^ {n} b_{i} * i^{x}\),最后答案就是\(f(k)\),考虑合并两颗子树\(t, p\),首先对于左子树\(t\)来说\(rk\)是不变的,直接继承,对于右子树来说,排名相当于偏移了一个\(sz(t)\),所以

\[f(rt) = f(t) + f(p) \]

\[f(rt) = f(t) + \sum b_{i} \times (i + sz(t)) ^ {p} \]

  • 二项式定理展开一下

\[\sum \sum\limits_{x = 0} ^ {p} b_{i} \times i ^ {x} \binom{n}{x} sz(t)^{p - x} \]

  • 所以就是的原来的\(f(p)\)再乘上一个\(\binom{n}{x} sz(t)^{p - x}\)就行。

  • 对于后边的被减项,因为要取模,我们设\(cnt[i][j]\)表示对于\(b_{i} mod \ w = i\),\(i \ mod \ w = j\)的数的个数,这个暴力转移就好了。

here




#include <bits/stdc++.h>

#define LL long long
#define Re register int
#define LD double
#define mes(x, y) memset(x, y, sizeof(x))
#define cpt(x, y) memcpy(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; ++ x)
#define fp(x, y, z)for(Re x = y; x >= z; -- x)
#define delfr(x, y, z)for(Re x = y; x < z; ++ x)
#define delfp(x, y, z)for(Re x = y; x > z; -- x)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define mk(x, y) make_pair(x, y)
#define pb(x) push_back(x)
#define re(x) return x
#define sec second
#define fst first
using namespace std;

namespace kiritokazuto{
    // auto read = [](){
    //     LL x = 0;
    //     int f = 1;
    //     char c;
    //     while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
    //     do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
    //     return x * f;
    // };
    inline char getc(){
        static char buf[1<<21],*p1,*p2;
        if(p1==p2){p1=buf,p2=buf+fread(buf,1,1<<18,stdin);if(p1==p2)return EOF;}
        return *p1++;
    }
    inline LL read(){
        LL x=0;char ch=getc();
        while(!isdigit(ch))ch=getc();
        while(isdigit(ch))x=(x<<1)+(x<<3)+(ch^48),ch=getc();
        return x;
    }
    template <typename T> fuc(void, write) (T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10);putchar(x % 10 | '0');
    }

}
using namespace kiritokazuto;
const int maxn = 5e6 + 10000, Inf = 2147483647, Mod = 998244353, lim = 100000;

fuc(LL, qpow)(LL a, LL b, LL mod) {
    LL res = 1;
    while (b) {
        if (b & 1)res = 1ll * (res * a) % mod;
        a = 1ll * (a * a) % mod;
        b >>= 1;
    }
    return res % mod;
}
LL C[6][6];
int a[maxn], K, w;
int n;
int Q;
int inv;
struct Tree {
    int cnt[6][6];
    int f[6];
    int sz;
    #define f(rt, x) tr[rt].f[x]
    #define cnt(rt, x, y) tr[rt].cnt[x][y]
    #define sz(rt) tr[rt].sz
}tr[maxn];
struct Seg_Tree {
    #define lsp(rt) (rt << 1)
    #define rsp(rt) (rt << 1 | 1)
    #define mid ((l + r) >>1)
    fuc(void, pushup)(int rt) {
        sz(rt) = sz(lsp(rt)) + sz(rsp(rt));
        fr(i, 0, K) {
            f(rt, i) = f(lsp(rt), i);
            fr(j, 0, i) {
                f(rt, i) = (f(rt, i) + f(rsp(rt), j) * C[i][j] % Mod * qpow(sz(lsp(rt)), i - j, Mod) % Mod) % Mod ;
            }
        }
        delfr(i, 0, w) delfr(j, 0, w) cnt(rt, i, j) = cnt(lsp(rt), i, j);
        delfr(i, 0, w) delfr(j, 0, w) (cnt(rt, i, (j + sz(lsp(rt))) % w)) += cnt(rsp(rt), i, j) %= Mod; 

    }
    fuc(void, insert)(int rt, int l, int r, int pos, int val) {
        if(l == r) {
            sz(rt) += val;//先加值
            fr(i, 0, K) {
                f(rt, i) += val * l * qpow(sz(rt) + (val == -1), i, Mod) % Mod;
                //val 为 1 或者 -1, 如果为-1,因为我这里的sz是先减过的,所以为了减掉一个贡献,我用的是原来没减的,所以把一补上(rk要原来的)
            }
            cnt(rt, l % w, ((sz(rt) + (val == -1)) % w)) += val;
            return;
        }
        if(pos <= mid)insert(lsp(rt), l, mid, pos, val);
        if(pos > mid)insert(rsp(rt), mid + 1, r, pos, val);
        pushup(rt);
    }
}T;
signed main() {
    fr(i, 0, 5)C[i][0] = 1;
    fr(i, 1, 5) fr(j, 1, i) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Mod;
    n = read(), K = read(), w = read();
    fr(i, 1, n)a[i] = read(), T.insert(1, 0, lim, a[i], 1);
    Q = read();

    inv = qpow(w, Mod- 2, Mod) % Mod;
    fr(i ,1, Q) {
        int x = read(), y = read();
        T.insert(1, 0, lim, a[x], -1);
        a[x] = y;
        T.insert(1, 0, lim, a[x], 1);
        LL ans = 0;
        delfr(i, 0, w) delfr(j, 0, w) {
            ans = (ans + (i * qpow(j, K, w) % w) % Mod * cnt(1, i, j) % Mod) % Mod;
        }
        write(((f(1, K) - ans + Mod) % Mod + Mod) % Mod * inv % Mod);
        ki;
    }
    return 0;
        
    
}

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标签:rt,No,int,18,LL,Rest,read,Mod,define
From: https://www.cnblogs.com/kiritokazuto/p/16787285.html

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