B. Skills
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
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.
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.
Examples
input
Copy
3 5 10 1 5 1 3 1
output
Copy
12 2 5 2
input
Copy
3 5 10 1 339 1 3 1
output
Copy
35 5 5 5
Note
In the first test the optimal strategy is to increase the second skill to its maximum, and increase the two others by 1.
In the second test one should increase all skills to maximum.
解题思路：先对原序列排序，然后从后往前枚举填满的个数。然后对于前面的进行二分，看最大能把最小值填到多大。然后更新答案即可。这里很多细节要注意！
#include <bits/stdc++.h>
using namespace std;
const int MAXN=100005;
typedef long long ll;
pair<ll,int> a[MAXN];
bool cmp(pair<ll,int> &a,pair<ll,int> &b){
return a.second<b.second;
}
ll need[MAXN];//把前i-1个填到跟第i个相同，需要多少次
ll sum[MAXN];//把前i个填到A，需要多少次
int main(){
int N;
ll A,CF,CM,M;
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);
int num=0;
int yuan=N;
while(a[N].first==A)
{
num++;
N--;
}
for(int i=1;i<=N;i++){
need[i]=need[i-1]+(a[i].first-a[i-1].first)*(i-1);
sum[i]=sum[i-1]+a[i].first;
}
ll ans=0;
int ansindex=0;
for(int i=0;i<=N;i++)//枚举填满的个数
{
ll tmp=M;
ll res=CF*(num+i);
tmp-=i*A-(sum[N]-sum[N-i]);//直接计算要多少次
if(tmp<0)
break;
if(i==N)//特殊处理
{
res+=CM*A;
if(res>ans){
ans=res;
ansindex=i;
}
break;
}
int l=1,r=N-i+1;//边界要注意细节
while(l<r){
int m=(l+r)/2;
if(need[m]<=tmp)
l=m+1;
else
r=m;
}
int index=l-1;
if(index==0)
index=1;
ll minn=a[index].first;
tmp-=need[index];
minn+=tmp/index;//计算最小能去到多大，
if(minn>A)
minn=A;
res+=minn*CM;
if(res>ans){
ans=res;
ansindex=i;
}
}
//把答案计算出来
ll tmp=M;
for(int j=N;j>=N-ansindex+1;j--){
tmp-=A-a[j].first;
a[j].first=A;
}
int l=1,r=N-ansindex+1;
while(l<r){
int m=(l+r)/2;
if(need[m]<=tmp)
l=m+1;
else
r=m;
}
int index=l-1;
if(index==0)
index=1;
tmp-=need[index];
ll num1=tmp/index;
for(int i=1;i<=index;i++){
a[i].first=a[index].first+num1;
a[i].first=min(A,a[i].first);
}
ll mod=tmp%index;
for(int i=1;i<=mod;i++){
a[i].first++;
a[i].first=min(A,a[i].first);
}
sort(a+1,a+yuan+1,cmp);
cout<<ans<<endl;
for(int i=1;i<=yuan;i++)
cout<<a[i].first<<" ";
cout<<endl;
return 0;
}