codeforces 1098E Fedya the Potter

这题思路很清楚。
应该就是先把b数组搞出来，然后再根据b算c数组的中位数。
首先先把b搞出来，如果用倍增，那么预处理(O(n log n log C))，算出b复杂度为(O(n log n log C log C))，稍微用一下gcd的trick就可以变成预处理(O(n (log n+log C))) ，算b复杂度(O(n log n log C))
如果用线段树加trick维护 预处理(O(n log C))，算b在树上二分也是(O(n log n log C))的。

然而我用了裸的线段树，所以是三个log的。
然后比较恶心的是类欧的部分。
假设要求
(sum_{i=0}^n lfloor frac{ai+b}{c} floor)
(sum_{i=0}^n sum_{j=1} [lfloor frac{ai+b}{c} floor geq j])
(sum_{i=0}^n sum_{j=1} [frac{ai+b}{c} geq j])
(sum_{i=0}^n sum_{j=0} [frac{ai+b}{c} geq j+1])
(sum_{i=0}^n sum_{j=0} [ai geq j*c+c-b])
(sum_{j=0} sum_{i=0}^n [i geq frac{j*c+c-b}{a}])
(sum_{j=0} sum_{i=0}^n [i > frac{j*c+c-b-1}{a}])
(sum_{j=0} n- lfloor frac{j*c+c-b-1}{a} floor)
令(m=lfloor frac{an+b}{c} floor)
(n*m-sum_{j=0} lfloor frac{j*c+c-b-1}{a} floor)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> p;
typedef long long ll;
namespace Segment{
	//also bz
	const int P=140000;
	int T[P];
	void update(int x,int l,int r,int pos,int v){
		if (l==r){
			T[x]=v;
			return;
		}
		int mid=(l+r)>>1;
		if (pos<=mid) update(x<<1,l,mid,pos,v);
		else update(x<<1|1,mid+1,r,pos,v);
		
		T[x]=__gcd(T[x<<1],T[x<<1|1]);
	}
	int ask(int x,int l,int r,int ll,int rr){
		if (ll<=l&&r<=rr) return T[x];
		int mid=(l+r)>>1,ret=0;
		if (ll<=mid) ret=__gcd(ret,ask(x<<1,l,mid,ll,rr));
		if (mid<rr) ret=__gcd(ret,ask(x<<1|1,mid+1,r,ll,rr));
		return ret;
	}
}

namespace Math{
	ll stand(ll lim1,ll a,ll b,ll c){
		//cerr<<"stand"<<lim1<<" "<<a<<" "<<b<<" "<<c<<endl;
		// x[0,lim1] ax+b/c
		if (a<0){
			ll pl=-(((a+1)/c)-1);
			return stand(lim1,a+pl*c,b,c)-pl*lim1*(lim1+1)/2;
		}
		if (a==0) return (lim1+1)*(b/c);
		if (a>=c||b>=c){
			return stand(lim1,a%c,b%c,c)+(a/c)*lim1*(lim1+1)/2+(b/c)*(lim1+1);
		}
		ll m=(a*lim1+b)/c;
		return lim1*m-stand(m-1,c,c-b-1,a);
	}
	ll divide(ll x,ll y){
		if (x>=0) return x/y;
		return x%y?x/y-1:x/y;
	}
	ll oc(ll a,ll b,ll lim1,ll lim2,ll c){
		//cerr<<"oc"<<a<<" "<<b<<" "<<lim1<<" "<<lim2<<" "<<c<<endl;
		if (lim1<0||lim2<0||c<0) return 0;
		//ax+by<=c  x [0,lim1] y [0,lim2]
		lim1=min(lim1,c/a);//max 0
		ll ret=lim1+1;//y=0
		//cerr<<"ret"<<ret<<endl;
		ll cd=min(max(divide(c-b*lim2,a),-1ll),lim1);
		/*for (int l=0,r=lim1,mid=(l+r)>>1; l<=r; mid=(l+r)>>1)
		  if ((c-a*mid)/b>=lim2) cd=mid,l=mid+1; else r=mid-1;*/
		ret+=(cd+1)*lim2;
		//cout<<"ret"<<ret<<" "<<cd<<endl;
		ret+=stand(lim1,-a,c,b);
		if (cd>-1) ret-=stand(cd,-a,c,b);
		//cerr<<"ret"<<ret<<endl;
		return ret;
	}
	ll co(ll a,ll b,ll c){
		//cerr<<"CP"<<a<<" "<<b<<" "<<c<<endl;
		ll num=b/a;
		if (num>c) return c*(c+1)/2;
		return (c-num)*num+num*(num+1)/2;
	}
}
const int N=50010;
const int C=100000;
ll b[C+10],sum[C+10],sb[C+10];
int pre[C+10];
ll calc(ll lim){
	//cerr<<"calc"<<lim<<endl;
	ll ret=0;
	for (int i=1; i<=C; ++i) if (b[i]) ret+=Math::co(i,lim-1,b[i]);
	for (int l=1,r=1; l<C; ++l){
		if (!b[l]) continue;
		if (l<r) ret+=b[l]*(sb[r-1]-sb[l]);
		for (; r<=C&&sum[r]-sum[l]<lim; ++r)
			if (l!=r&&b[r]) ret+=Math::oc(l,r,b[l]-1,b[r]-1,lim-1-(sum[r-1]-sum[l])-l-r);
		if (r<=C){
			if (l!=r&&b[r]) ret+=Math::oc(l,r,b[l]-1,b[r]-1,lim-1-(sum[r-1]-sum[l])-l-r);
		}
	}
	return ret;
}
int n;
int main(){
	//cerr<<Math::oc(2,5,0,1,4)<<endl;
	//return 0;
	scanf("%d",&n);
	for (int i=1,x; i<=n; ++i){
		scanf("%d",&x);
		int la=i;
		Segment::update(1,1,n,i,x);
		while (la){
			int tmp=la;
			for (int l=1,r=la,mid=(l+r)>>1; l<=r; mid=(l+r)>>1)
				if (Segment::ask(1,1,n,mid,i)==Segment::ask(1,1,n,la,i)) la=mid,r=mid-1; else l=mid+1;
			b[Segment::ask(1,1,n,la,i)]+=tmp-la+1;
			--la;
		}
	}
	//for (int i=1; i<=20; ++i) cout<<b[i]<<" "; cout<<endl;
	for (int i=1; i<=C; ++i){
		sb[i]=sb[i-1]+b[i];
		sum[i]=sum[i-1]+b[i]*i;
	}
	int la=0;
	for (int i=1; i<=C; ++i)
		if (b[i]){
			pre[i]=la;
			la=i;
		}
	ll bat=(ll)n*(n+1)/2,ans;
	bat=bat*(bat+1)/2;
	//cerr<<"BAT"<<bat<<endl;
   	//cerr<<calc(8)<<endl;
	//return 0;
	for (ll l=1,r=(ll)C*n*(n+1)/2,mid=(l+r)>>1; l<=r; mid=(l+r)>>1){
		//cerr<<"NA"<<ans<<endl;
		if (calc(mid)+1<=(bat+1)/2) ans=mid,l=mid+1;
		else r=mid-1;
	}
	cout<<ans;
}