题目链接：
2 seconds
256 megabytes
standard input
standard output
Lesha plays the recently published new version of the legendary game hacknet. In this version character skill mechanism was introduced. Now, each player character has exactly n skills. Each skill is represented by a non-negative integer ai — the current skill level. All skills have the same maximum level A.
Along with the skills, global ranking of all players was added. Players are ranked according to the so-called Force. The Force of a player is the sum of the following values:
- The number of skills that a character has perfected (i.e., such that ai = A), multiplied by coefficient cf.
- The minimum skill level among all skills (min ai), multiplied by coefficient cm.
Now Lesha has m hacknetian currency units, which he is willing to spend. Each currency unit can increase the current level of any skill by 1 (if it's not equal to A yet). Help him spend his money in order to achieve the maximum possible value of the Force.
The first line of the input contains five space-separated integers n, A, cf, cm and m (1 ≤ n ≤ 100 000, 1 ≤ A ≤ 109, 0 ≤ cf, cm ≤ 1000,0 ≤ m ≤ 1015).
The second line contains exactly n integers ai (0 ≤ ai ≤ A), separated by spaces, — the current levels of skills.
On the first line print the maximum value of the Force that the character can achieve using no more than m currency units.
On the second line print n integers a'i (ai ≤ a'i ≤ A), skill levels which one must achieve in order to reach the specified value of the Force, while using no more than m currency units. Numbers should be separated by spaces.
3 5 10 1 5
1 3 1
12
2 5 2
3 5 10 1 339
1 3 1
35
5 5 5 题意： m次操作,每次可以任选一个a[i]++,最后要求min(a[i])*cm+num(a[i]==A)*cf最大; 思路： 先排序,再枚举a[i]==A的个数,然后二分min(a[i]),根据贪心的原则,要使num尽量多,那么应该从最大的a[i]开始变成A,要使min(a[i])尽可能大,那么应该把最小的++; AC代码：
//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio> using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());
for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());
F && (num=-num);
}
int stk[], tp;
template<class T> inline void print(T p) {
if(!p) { puts(""); return; }
while(p) stk[++ tp] = p%, p/=;
while(tp) putchar(stk[tp--] + '');
putchar('\n');
} const LL mod=1e9+;
const double PI=acos(-1.0);
const LL inf=1e14;
const int N=1e5+; int n,cf,cm;
LL m,pre[N],nex[N],A;
struct node
{
int a,id;
}po[N];
int cmp(node x,node y)
{
return x.a<y.a;
}
int cmp1(node x,node y)
{
return x.id<y.id;
}
int check(LL y,LL x,int fr)
{
int l=,r=fr;
while(l<=r)
{
int mid=(l+r)>>;
if(po[mid].a<=x)l=mid+;
else r=mid-;
}
if(y>=x*(l-)-pre[l-])return ;
return ;
} int fipos,fnum;
int tpos,tnum;
LL solve(int x)
{
int num=n+-x;
if(m>=num*A-nex[x])
{
LL temp=m-num*A+nex[x];
LL l=po[].a,r=A;
while(l<=r)
{
LL mid=(l+r)>>;
if(check(temp,mid,x-))l=mid+;
else r=mid-;
}
tpos=(int)(l-),tnum=num;
return (l-)*cm+num*cf;
}
return ;
}
int main()
{
read(n);read(A);read(cf);read(cm);read(m);
Riep(n)read(po[i].a),po[i].id=i;
sort(po+,po+n+,cmp);
pre[]=;nex[n+]=;
for(int i=;i<=n;i++)pre[i]=pre[i-]+po[i].a;
for(int i=n;i>;i--)nex[i]=nex[i+]+po[i].a;
int pos=;
LL ans=;
for(int i=n+;i>;i--)
{
LL d=solve(i);
if(d>ans)
{
ans=d;
pos=i;
fipos=tpos;
fnum=tnum;
}
}
cout<<ans<<"\n";
for(int i=n;i>n-fnum;i--)
{
po[i].a=(int)A;
}
for(int i=;i<=n;i++)
{
if(po[i].a<fipos)po[i].a=fipos;
}
sort(po+,po+n+,cmp1);
Riep(n)printf("%d ",po[i].a);
return ;
}