解题：ZJOI 2013 K大数查询

题面

树套树，权值线段树套序列线段树，每次在在权值线段树上的每棵子树上做区间加，查询的时候左右子树二分

本来想两个都动态开点的，这样能体现树套树在线的优越性。但是常数太大惹，所以外层直接固定建树了QAQ

就是不写整体二分╭(╯^╰)╮

Warning：Extremely ugly code detected

 // luogu-judger-enable-o2
 //I CAN PASS THE TEST
 #include<cstdio>
 #include<cctype>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=;
 int n,m,f,op,tot;
 long long val[N];
 int root[N],lson[N],rson[N],laz[N];
 void read1(int &x)
 {
     x=; char ch=getchar();
     while(!isdigit(ch))
         ch=getchar();
     while(isdigit(ch))
         x=(x<<)+(x<<)+(ch^),ch=getchar();
 }
 void read2(int &x)
 {
     x=f=; char ch=getchar();
     while(!isdigit(ch))
         f|=ch=='-',ch=getchar();
     while(isdigit(ch))
         x=(x<<)+(x<<)+(ch^),ch=getchar();
     x=f?-x:x;
 }
 void read3(long long &x)
 {
     x=; char ch=getchar();
     while(!isdigit(ch))
         ch=getchar();
     while(isdigit(ch))
         x=(x<<)+(x<<)+(ch^),ch=getchar();
 }
 void write(int x)
 {
     if(x>) write(x/);
     putchar(x%|);
 }
 void Insert(int &nde,int ll,int rr,int l,int r)
 {
     if(ll>r||rr<l) return ;
     if(!nde) nde=++tot;
     val[nde]+=min(r,rr)-max(l,ll)+;
     if(ll>=l&&rr<=r)
            ++laz[nde];
     else
     {
         int mid=(ll+rr)>>;
         Insert(lson[nde],ll,mid,l,r);
         Insert(rson[nde],mid+,rr,l,r);
     }
 }
 void Add(int l,int r,int v)
 {
     int nde=,ll=,rr=*n;
     while(true)
     {
         Insert(root[nde],,n,l,r);
         if(ll^rr)
         {
             int mid=(ll+rr)>>;
             (v<=mid)?(nde<<=,rr=mid):(nde<<=,nde|=,ll=mid+);
         }
         else return ;
     }
 }
 long long Query(int nde,int ll,int rr,int l,int r,int lz)
 {
     if(ll>r||rr<l)
         return ;
     else if(ll>=l&&rr<=r)
         return val[nde]+1ll*lz*(rr-ll+);
     else
     {
         int mid=(ll+rr)>>;
         return Query(lson[nde],ll,mid,l,r,lz+laz[nde])+Query(rson[nde],mid+,rr,l,r,lz+laz[nde]);
     }
 }
 int main()
 {
     register int i;
     read1(n),read1(m);
     for(i=;i<=m;i++)
     {
         read1(op);
         if(op==)
         {
             int t1,t2,t3;
             read1(t1),read1(t2),read2(t3);
             int nde=,ll=,rr=*n; t3=n-t3+;
             while(true)
             {
                 Insert(root[nde],,n,t1,t2);
                 if(!(ll^rr)) break;
                 int mid=(ll+rr)>>;
                 (t3<=mid)?(nde<<=,rr=mid):(nde<<=,nde|=,ll=mid+);
             }
         }
         else
         {
             int t1,t2; long long t4;
             read1(t1),read1(t2),read3(t4);
             int nde=,ll=,rr=*n;
             while(true)
             {
                 if(ll==rr) {write(n-ll+),puts(""); break;}
                 int mid=(ll+rr)/;
                 long long qry=Query(root[nde<<],,n,t1,t2,);
                 (t4<=qry)?(nde<<=,rr=mid):(nde<<=,nde|=,ll=mid+,t4-=qry);
             }
         }
     }
     return ;
 }