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思路

良心来说这题还挺思维的

我没看题解也不知道要这样维护

把每次斩断的点分别放进两个队列里面

因为要维护增长，所以可以让新进队的节点来一个负增长？

是不是就好了？

然后很容易发现因为在原始队列中小的数比大的数如果多增加了\(k \times q\)在剩下两个队列中大的分别比小的多增加大于等于\(k \times q\)

然后随便维护一下

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//Author: dream_maker#include
    
     using namespace std;//----------------------------------------------//typenametypedef long long ll;//convenient for#define fu(a, b, c) for (int a = b; a <= c; ++a)#define fd(a, b, c) for (int a = b; a >= c; --a)#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)//inf of different typenameconst int INF_of_int = 1e9;const ll INF_of_ll = 1e18;//fast read and writetemplate 
     
      void Read(T &x) {  bool w = 1;x = 0;  char c = getchar();  while (!isdigit(c) && c != '-') c = getchar();  if (c == '-') w = 0, c = getchar();  while (isdigit(c)) {    x = (x<<1) + (x<<3) + c -'0';    c = getchar();  }  if (!w) x = -x;}template 
      
       void Write(T x) {  if (x < 0) {    putchar('-');    x = -x;   }  if (x > 9) Write(x / 10);  putchar(x % 10 + '0');}//----------------------------------------------const int N = 1000010;int n, m;ll q, u, v, t, a[N];queue
       
         Q[4];ll find_pos() {  int res = 0;  if (Q[1].size() && (!res || Q[1].front() > Q[res].front())) res = 1;  if (Q[2].size() && (!res || Q[2].front() > Q[res].front())) res = 2;  if (Q[3].size() && (!res || Q[3].front() > Q[res].front())) res = 3;  return res;}int main() {  Read(n), Read(m), Read(q), Read(u), Read(v), Read(t);  fu(i, 1, n) Read(a[i]);  sort(a + 1, a + n + 1);  fd(i, n, 1) Q[1].push(a[i]);  fu(i, 1, m) {    ll pos = find_pos(), len = Q[pos].front();    Q[pos].pop();    len += (i - 1) * q;    if (i % t == 0) {      Write(len);      putchar(' ');    }    ll l = len * u / v, r = len - l;    l -= i * q; r -= i * q;    Q[2].push(l);    Q[3].push(r);  }  putchar('\n');  fu(i, 1, n + m) {    ll pos = find_pos(), len = Q[pos].front();    Q[pos].pop();    len += m * q;    if (i % t == 0) {      Write(len);      putchar(' ');    }  }  return 0;}