CF 830C (Bamboo partition-satisfies the maximum value of D of \sum_{i=1}^n{D\lceil ai/d\rceil-a_i} \leq k) __ Number theory

Ask: Find the largest positive integer d, so that ∑ni=1d⌈ai/d⌉−ai≤k \sum_{i=1}^n{D\lceil ai/d\rceil-a_i} \leq K
observed as long as ⌈ai/d⌉ (i=1⋯n) \lceil AI/D\RC The Eil (i=1\cdots N) is invariant, the function is linear, and the
therefore solves each segment separately.
Segment Point total Nmax (AI) −−−−−−−√ n\sqrt {max (a_i)}

#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) to (int i=n;i>=k;i--) #define REPD (i,n) for (int i=n;i& gt;=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]) #defin
E 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 #define SE sec Ond #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). The Begin (), (x). End () Ty
Pedef 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 (a) ll n,a[maxn],k;
    LL Calc (ll d) {ll =;
    for (i,n) {p+= (a[i]%d+d)%d;
} return p;
    BOOL Check (ll D) {if (d<0) return 0;
    ll p =;
    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];p re=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;
        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;
 }