题意：求最大的正整数d，，使∑ni=1d⌈ai/d⌉−ai≤k
观察发现只要⌈ai/d⌉(i=1⋯n)不变, 函数是线性的，
因此对每段分别求解。
分段点共nmax(ai)−−−−−−−√个
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (110)
ll n,a[MAXN],k;
ll calc(ll d) {
ll p=0;
For(i,n) {
p+=(a[i]%d+d)%d;
}
return p;
}
bool check(ll d) {
if (d<0) return 0;
ll p=0;
For(i,n) {
p+=(a[i]%d+d)%d;
}
// cout<<d<<' '<<p<<' '<<endl;
return p<=k;
}
ll work(ll l,ll r) {
ll ans=-1;
if (l>r) return ans;
if (l==r) {
if (check(l)) ans=l;
return ans;
}
ll p=calc(l),delta=calc(l+1)-p;
ll mm;
if(!delta) {
mm=r;
}
else {
ll c=(k-p)/(delta);
mm=min(l+c,r);
}
if (check(mm)) ans=mm;
return ans;
}
vector<ll> S;
int main()
{
// freopen("C.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>k;
For(i,n) a[i]=read();
sort(a+1,a+1+n);
For(i,n) a[i]=-a[i];
ll ans=0,nxt;
For(i,n) {
ll nxt=1;
for(ll pre=1;pre<=-a[i];pre=nxt+1) {
nxt=a[i]/(a[i]/pre);
S.pb(nxt);
}
}
S.pb(1);
sort(ALL(S));
S.erase(unique(ALL(S)),S.end());
vector<ll>::reverse_iterator it;
work(35,36);
ll pre=*S.rbegin();
for(it=S.rbegin();it!=S.rend();it++) {
if (check(pre)) {
ans=pre; break;
}
ll nxt=*it;
ll p=work(nxt+1,pre);
if (p!=-1) {
ans=max(ans,p); break;
}
pre=nxt;
}
if (check(1)) ans=max(ans,1LL);
ll tot=-a[n]*n;
For(i,n) tot-=-a[i];
if(tot<=k) {
ll p=(k-tot)/n;
if(check(-a[n]+p)) ans=max(ans,-a[n]+p);
}
cout<<ans<<endl;
return 0;
}
