数据较小（$n\le 12$），所以本蒟蒻用搜索写的，勿喷

#include<bits/stdc++.h>
using namespace std;
int n,x,y,z,ans=1e9,a[110],dist[20][1005];
void dfs(int t,int u,int s)
{
	if(s>=ans)return;
	if(t==n+1)
	{
		ans=s;
		return;
	}
	if(u<=a[t])//比至少要的员工少
	{
		if(dist[t][a[t]]>s+(a[t]-u)*x+a[t]*y)//剪枝
		{
			dist[t][a[t]]=s+(a[t]-u)*x+a[t]*y;
			dfs(t+1,a[t],s+(a[t]-u)*x+a[t]*y);
		}
	}
	else
	{
		for(int i=0;i<=u-a[t];i++)//可以裁0~当前员工-至少员工个
			if(dist[t][u-i]>s+i*z+(u-i)*y)//剪枝
			{
				dist[t][u-i]=s+i*z+(u-i)*y;
				dfs(t+1,u-i,s+i*z+(u-i)*y);
			}
	}
}
int main()
{
	memset(dist,0x3f,sizeof dist);
	cin>>n>>x>>y>>z;
	for(int i=1;i<=n;i++)cin>>a[i];
	dfs(1,0,0);//当前月数 员工人数 花钱费用
	cout<<ans;
	return 0;
}

不点赞是$\fcolorbox{gray}{gray}{\color{gray}{gay}}$!!!

#include <bits/stdc++.h>
using namespace std;
int n, hire, paid, cut, w[20], had, cost;
int main() {
	cin >> n >> hire >> paid >> cut;
	for (int i = 1; i <= n; ++ i) cin >> w[i];
	had = w[1], cost = (hire + paid) * w[1];
	for (int i = 2; i <= n; ++ i) {
		cost += paid * w[i];
		if (w[i] >= had) {
			cost += hire * (w[i] - had);
			had = w[i];
			continue;
		}
		int gap = had - w[i];
		for (int j = 1; j <= gap; ++ j) {
			int sub = n + 1, cost1 = 0, cost2 = 0;
			for (int k = i + 1; k <= n; ++ k)
				if (w[k] >= w[i] + j) {
					sub = k;
					cost1 = hire;
					break;
				}
			cost1 += cut, cost2 = paid * (sub - i);
			if (cost1 >= cost2) cost += paid;
				else cost += cut, -- had;
		}
	}
	cout << cost;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n,a,b,c,ned[200],ans,m,c1,c2,k;
int main()
{
	cin>>n;
	cin>>a>>b>>c;
	for(int i=1;i<=n;i++) cin>>ned[i];
	ans=(a+b)*ned[1];
	for(int i=2;i<=n;i++)
	{
		ans=ans+b*ned[i],m=ned[i]-ned[i-1];
		if(m>=0) ans=ans+a*m;
		else
		{
			m=abs(m);
			for(int j=1;j<=m;j++)
			{
				c1=0;
				for(k=i+1;k<=n;k++)
					if(ned[k]>ned[i])
					{
						c1=a;
						break;
					}
				c1+=c;
				c2=b*(k-i);
				if(c1>=c2) ned[i]++,ans+=b;
					else 
					{
						ans+=c*(m-j+1);
						break;
					}
			}
		}
	}
	cout<<ans;
}

刀片解，不喜勿喷

$dp_{ij}$：第$i$个月雇佣$j$个人的最小花费。 $dp_{ij}$通过查找前一个月雇佣人的最小金额来求最小值得出。

#include<bits/stdc++.h>
using namespace std;
int dp[15][1005],n,x,y,z,a[10005],mi=INT_MAX;
int main(){
cin>>n>>x>>y>>z;
cin>>a[1];
for(int i=a[1];i<=1000;i++)dp[1][i]=i*(x+y);//初始化
for(int i=2;i<=n;i++){
cin>>a[i];//输入
for(int j=a[i];j<=1000;j++){
dp[i][j]=INT_MAX;//初始化
for(int k=a[i-1];k<=1000;k++){
int s;
if(k>j)s=(k-j)*z+dp[i-1][k]+y*j;//若人多了，就解雇几个
else s=(j-k)*x+dp[i-1][k]+y*j;//若人少了，就雇佣几个
dp[i][j]=min(s,dp[i][j]);//求最小值
}
if(i==n)mi=min(mi,dp[i][j]);//求答案
}
}cout<<mi;//输出
return O;
}

刀片解，不喜勿喷

#include<bits/stdc++.h>
using namespace std;
int dp[15][1005],n,x,y,z,a[10005],mi=INT_MAX;
int main(){
cin>>n>>x>>y>>z;
cin>>a[1];
for(int i=a[1];i<=1000;i++)dp[1][i]=i*(x+y);//初始化
for(int i=2;i<=n;i++){
cin>>a[i];//输入
for(int j=a[i];j<=1000;j++){
dp[i][j]=INT_MAX;//初始化
for(int k=a[i-1];k<=1000;k++){
int s;
if(k>j)s=(k-j)*z+dp[i-1][k]+y*j;
else s=(j-k)*x+dp[i-1][k]+y*j;
dp[i][j]=min(s,dp[i][j]);//求最小值
}
if(i==n)mi=min(mi,dp[i][j]);//求答案
}
}cout<<mi;//输出
return O;
}