The meaning of problems
Each insertion interval \ ([L_i, R_i] \ ) number between query median.
analysis
Into the range of discrete points, you can use the tree line to support the update and query.
We will right the interval +1 point, the length of the interval is a right end point minus the left point, then we will look at all endpoints with discrete segment tree maintenance on the line.
For example the interval \ ([1,2], [2,4] \)
+1 becomes the right end of the interval \ ([1,3), [2,5) \)
After discrete \ (1,2,3,4 \)
\ (1 \) representative of the interval \ ([1,2) \) , \ (2 \) representative of the interval \ ([2,3) \) , \ (3 \) representative of the interval \ ([3,5) \) , \ (4 \) does not represent any range.
- Update interval \ ([l, r) \ ) when (r + 1 plus the right end point), since the point \ (r-1 \) section is represented by \ ([R & lt-. 1, R & lt) \) , we update interval \ ([l, r-1 ] \) of points on it.
- Find the point where the median interval from a tree half a query, and then calculate the interval in which the number is on the line.
Code
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define lson l,mid,p<<1
#define rson mid+1,r,p<<1|1
#define ll long long
using namespace std;
const int inf=1e9;
const int mod=1e9+7;
const int maxn=8e5+10;
const int N=1e9;
int n,m;
ll X1,X2,A1,B1,C1,M1,Y1,Y2,A2,B2,C2,M2;
int L[maxn],R[maxn],b[maxn];
ll a[maxn<<2],tag[maxn<<2];
void up(int dl,int dr,int l,int r,int p){
a[p]+=b[dr+1]-b[dl];
if(l==dl&&r==dr) return tag[p]++,void();
int mid=l+r>>1;
if(dr<=mid) up(dl,dr,lson);
else if(dl>mid) up(dl,dr,rson);
else up(dl,mid,lson),up(mid+1,dr,rson);
}
int qy(int l,int r,int p,ll x,ll k){
if(l==r){
ll g=a[p]/(b[l+1]-b[l])+k;
return b[l]+(x+g-1)/g-1;
}
int mid=l+r>>1;
ll res=a[p<<1]+(k+tag[p])*(b[mid+1]-b[l]);
if(x<=res) return qy(lson,x,k+tag[p]);
else return qy(rson,x-res,k+tag[p]);
}
int main(){
//ios::sync_with_stdio(false);
//freopen("in","r",stdin);
scanf("%d",&n);
cin>>X1>>X2>>A1>>B1>>C1>>M1;
cin>>Y1>>Y2>>A2>>B2>>C2>>M2;
ll X=X1,Y=Y1,K=0;
for(int i=1;i<=min(n,2);i++){
if(i==2) X=X2,Y=Y2;
L[i]=min(X,Y)+1,R[i]=max(X,Y)+1;
b[++m]=L[i];b[++m]=R[i]+1;
}
for(int i=3;i<=n;i++){
X=(A1*X2%M1+B1*X1%M1+C1)%M1;
Y=(A2*Y2%M2+B2*Y1%M2+C2)%M2;
L[i]=min(X,Y)+1,R[i]=max(X,Y)+1;
b[++m]=L[i];b[++m]=R[i]+1;
X1=X2;X2=X;
Y1=Y2;Y2=Y;
}
sort(b+1,b+m+1);
m=unique(b+1,b+m+1)-b-1;
for(int i=1;i<=n;i++){
L[i]=lower_bound(b+1,b+m+1,L[i])-b;
R[i]=lower_bound(b+1,b+m+1,R[i]+1)-b;
K+=b[R[i]]-b[L[i]];
up(L[i],R[i]-1,1,m,1);
printf("%d\n",qy(1,m,1,(K+1)/2,0));
}
return 0;
}