bzoj3110: [Zjoi2013]K大数查询（权值线段树套区间线段树）

题目
内层线段树维护权值k在[l,r]内出现次数

#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ll;
#define mid ((l+r)>>1)
const int N=200002,M=20000002;
int n,m,op,x,y,z,lz[M],L[M],R[M],cnt,rt[N];
ll sum[M];
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
    int x=0,fl=1;char ch=gc();
    for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
    for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
    return x*fl;
}
inline void wri(int a){if(a>=10)wri(a/10);putchar(a%10|48);}
inline void wln(int a){if(a<0)a=-a,putchar('-');wri(a),puts("");}
void down(int t,int l,int r){
    if (lz[t]){
        if (!L[t]) L[t]=++cnt;
        if (!R[t]) R[t]=++cnt;
        sum[L[t]]+=(mid-l+1)*lz[t];
        sum[R[t]]+=(r-mid)*lz[t];
        lz[L[t]]+=lz[t];
        lz[R[t]]+=lz[t];
        lz[t]=0;
    }
}
void update(int &t,int l,int r,int x,int y){
    if (!t) t=++cnt;
    if (x<=l && r<=y){
        sum[t]+=r-l+1;
        lz[t]++;
        return;
    }
    down(t,l,r);
    if (x<=mid) update(L[t],l,mid,x,y);
    if (mid<y) update(R[t],mid+1,r,x,y);
    sum[t]=sum[L[t]]+sum[R[t]];
}
ll query(int t,int l,int r,int x,int y){
    if (!t) return 0;
    if (x<=l && r<=y) return sum[t];
    down(t,l,r);
    ll ans=0;
    if (x<=mid) ans+=query(L[t],l,mid,x,y);
    if (mid<y) ans+=query(R[t],mid+1,r,x,y);
    return ans;
}
void ins(){
    int t=1,l=1,r=n;
    while (l<r){
        update(rt[t],1,n,x,y);
        if (z<=mid) r=mid,t<<=1;
        else l=mid+1,t=t<<1|1;
    }
    update(rt[t],1,n,x,y);
}
int solve(){
    int t=1,l=1,r=n;
    while (l<r){
        ll p=query(rt[t<<1],1,n,x,y);
        if (z<=p) r=mid,t<<=1;
        else l=mid+1,t=t<<1|1,z-=p;
    }
    return l;
}
int main(){
    n=rd(),m=rd();
    for (;m--;){
        op=rd(),x=rd(),y=rd(),z=rd();
        if (op==1) z=n+1-z,ins();
        else wln(n+1-solve());
    }
}