CF730F. Ber Patio

Solution

由于$\sum a_i$只有$10^5$，即除掉原有的$b$，代金券最多为$10^4$，因此我们令$f[i][j]$表示到了第$i$天，靠现金获得的代金券共为$j$的最小现金和。

转移时枚举第$i+1$天用$k$张代金券，其他用现金的方法支付：
$$upmin(f[i+1][j+(a[i+1]-k)/10],f[i][j]+(a[i+1]-k))$$
显然$k$最大不超过$min(\frac{a_i}{2},b-(s[i]-f[i][j])+j)$，能在$3s$内解决问题。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second
#define int llusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=300005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
vector<int> Ans;
int sum=0,f[2][10005],a[5005],s[5005],frm[5005][10005];
signed main()
{int n=read(),b=read();for (int i=1;i<=n;i++) s[i]=s[i-1]+(a[i]=read());for (int i=0;i<=1;i++)for (int j=0;j<=s[n]/10;j++) f[i][j]=INF;f[0][0]=0;int nw=0;for (int i=0;i<n;i++){nw^=1;for (int j=0;j<=s[i]/10;j++) f[nw][j]=INF;for (int j=0;j<=s[i]/10;j++)for (int k=0;k<=min(a[i+1]/2,b-(s[i]-f[nw^1][j])+j);k++) if (upmin(f[nw][j+(a[i+1]-k)/10],f[nw^1][j]+a[i+1]-k)) frm[i+1][j+(a[i+1]-k)/10]=k;}int ans=INF,Nw=0;for (int i=0;i<=s[n]/10;i++) if (upmin(ans,f[nw][i])) Nw=i;printf("%lld\n",ans);for (int i=n;i>=1;i--) Ans.PB(frm[i][Nw]),Nw=Nw-((a[i]-frm[i][Nw])/10);for (int i=n-1;i>=0;i--) printf("%lld ",Ans[i]);return 0;
}