Codeforces Round #339 (Div. 1) B. Skills 暴力 二分

B. Skills

题目连接：

http://www.codeforces.com/contest/613/problem/B

Description

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.

Input

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.

Output

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.

Sample Input

3 5 10 1 5

1 3 1

Sample Output

12

2 5 2

Hint

题意

你有n个技能，每个技能最高A级，你还有m个技能点没加

然后你的实力等于最低的技能等级*cm+等级加满的技能数量*cf

现在问你怎么加点，可以使得你的实力最大

题解：

首先贪心，我加满的技能，肯定是从高往低加

我要提高最低的技能，肯定从低到高加

那么我就枚举我加满的技能数量，然后二分我究竟能够加多少个最低的技能。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n;
long long A,cf,cm,m;
pair<long long ,int> a[maxn];
long long b[maxn],c[maxn];
bool cmp(pair<long long ,int> aa,pair<long long ,int> bb)
{
return aa.second<bb.second;
}
int main()
{
scanf("%d%lld%lld%lld%lld",&n,&A,&cf,&cm,&m);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i].first),a[i].second=i;
sort(a+1,a+1+n);
for(int i=1;i<=n;i++)
{
b[i]=b[i-1]+a[i].first;
c[i]=a[i].first*i-b[i];
}
long long ans = 0,ans1 = 0,ans2 = 0;
for(int i=0;i<n;i++)
{
if(m<0)break;
int pos = upper_bound(c,c+1+(n-i),m)-c-1;
long long q = (m-c[pos])/pos+a[pos].first;
q = min(q,A);
long long tmp = q*cm+i*cf;
if(tmp>ans)
{
ans = tmp;
ans1 = q,ans2 = i;
}
m = m - (A - a[n-i].first);
}
if(m>=0)
ans = A*cm+n*cf;
printf("%lld\n",ans);
for(int i=1;i<=n;i++)
{
if(n-i<ans2)
a[i].first = A;
else if(a[i].first<=ans1)
a[i].first = ans1;
}
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++)
printf("%lld ",a[i].first);
}