[P2827] problem solution earthworm

Solution earthworm

Subject to the effect: In the beginning there \ (n \) only earthworms, you operate \ (m \) times. Every time grasping the longest one, note length \ (the X-\) , cut into \ (\ lfloor PX \ rfloor \) , \ (X-\ lfloor PX \ rfloor \) two back, while in addition to these two All worms only increase the length of the outer \ (Q \) , each Q earthworms cut length, \ (m \) length after the operations of all the earthworms

queue

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First we solve collective increase in addition to the new \ (q \) of the problem, we can maintain a delta \ (Delta \) , indicates the length of all the earthworms have increased \ (Delta \) , two new earthworms without \ (Q \) , which we will length minus \ (Q \) then can be treated, so there may be negative in the process

It is not difficult to think of a simple analog rough idea, with a large root heap, while maintaining throw back after each increment to take the biggest cut \ (Delta \) . But this will take a \ (log \) , can not afford

The key is added after two earthworms than the original cut short , so every time we get out of the earthworm is decreasing length (without considering the incremental \ (Delta \) ), with the sequences into this topic and ideas, with a queue \ (Q_1 \) stores the original \ (n-\) earthworms (like row a sequence), \ (Q_2 \) storing cut \ (\ lfloor px \ rfloor \ ) earthworms, \ (Q_3 \) storing cut to \ (x- \ lfloor px \ rfloor \) earthworms, each taking from three queue maximum queue head, cut worms were thrown \ (Q_2 \) , \ (Q_3 \) maintenance \ (Delta \ ) to

#include <algorithm>
#include <cstdio>
#include <cctype>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
inline int read(){
    int x = 0;char c = getchar();
    while(!isdigit(c))c = getchar();
    while(isdigit(c))x = x * 10 + c - '0',c = getchar();
    return x;
}
queue<ll> Q[4];
int val[maxn],n,m,q,u,v,t;
ll delta;
inline int findmax(){
    int res = 0;
    for(int i = 1;i <= 3;i++)
        if(!Q[i].empty() && (!res || Q[i].front() > Q[res].front()))res = i;
    return res;
}
inline ll cut(ll x){return 1ll * u * x / v;}
inline void solve(int tim){
    ll x = Q[findmax()].front();Q[findmax()].pop();
    if(tim % t == 0)printf("%lld ",x + delta);
    Q[2].push(cut(x + delta) - delta - q);Q[3].push(x - cut(x + delta) - q);
    delta += q;
}
int main(){
    n = read(),m = read(),q = read(),u = read(),v = read(),t = read();
    for(int i = 1;i <= n;i++)val[i] = read();
    sort(val + 1,val + 1 + n);
    for(int i = n;i >= 1;i--)Q[1].push(val[i]);
    for(int i = 1;i <= m;i++)solve(i);
    putchar('\n');
    for(int i = 1;i * t <= n + m;i++){
        for(int j = 1;j <= t - 1;j++)
            Q[findmax()].pop();
        printf("%lld ",Q[findmax()].front() + delta);
        Q[findmax()].pop();
    }
    putchar('\n');
    return 0;
}