CF 830C(Bamboo Partition-满足\sum_{i=1}^n{ d\lceil ai/d\rceil-a_i } \leq k的d的最大值)

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;}