제출 #639013

#제출 시각아이디문제언어결과실행 시간메모리
639013zaneyuMonkey and Apple-trees (IZhO12_apple)C++14
100 / 100
399 ms174916 KiB
/*input
4
2 2 3
1 1 3
2 2 3
1 -1 3w
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET(v,x) lower_bound(ALL(v),x)-v.begin()
const int maxn=1e5+3;
const int INF=0x3f3f3f3f;
const int MOD=1e6+3;
const ld PI=acos(-1.0l);
const ld eps=1e-6;
const ll INF64=9e18+1;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"");
    return out;
}
int mult(int a,int b){
    return 1LL*a*b%MOD;
}
ll mypow(ll a,ll b,ll mod){
    if(b<=0) return 1;
    a%=mod;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}
struct node{
    ll sum=0,lazy=0;
    int lc=0,rc=0;
}seg[80*maxn];
//check this line
int tot=2;
inline void pushdown(int idx,int l,int r){
    if(!seg[idx].lazy) return;
    seg[idx].sum=r-l+1;
    if(l!=r){
        if(!seg[idx].lc) seg[idx].lc=(tot++);
        if(!seg[idx].rc) seg[idx].rc=(tot++);
        seg[seg[idx].lc].lazy=seg[idx].lazy;
        seg[seg[idx].rc].lazy=seg[idx].lazy;
    }
    seg[idx].lazy=0;
}
inline void mdf(int &idx,int l,int r,int ql,int qr,int x){
    if(!idx){
        idx=(tot++);
    }
    pushdown(idx,l,r);
    if(r<ql or l>qr) return;
    if(ql<=l and r<=qr){
        seg[idx].lazy=x;
        pushdown(idx,l,r);
        return;
    }
    int mid=(l+r)/2;
    mdf(seg[idx].lc,l,mid,ql,qr,x);
    mdf(seg[idx].rc,mid+1,r,ql,qr,x);
    seg[idx].sum=(seg[seg[idx].lc].sum+seg[seg[idx].rc].sum);
}
inline ll query(int &idx,int l,int r,int ql,int qr){
    if(r<ql or l>qr) return 0;
    if(!idx){
        idx=(tot++);
    }
    pushdown(idx,l,r);
    if(ql<=l and r<=qr){
        return seg[idx].sum;
    }
    int mid=(l+r)/2;
    return (query(seg[idx].lc,l,mid,ql,qr)+query(seg[idx].rc,mid+1,r,ql,qr));
}

int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int q;
    cin>>q;
    int c=0;
    int rt=1;
    int mx=1e9;
    while(q--){
        int d,x,y;
        cin>>d>>x>>y;
        x+=c,y+=c;
        if(d==2){
            mdf(rt,1,mx,x,y,1);
        }
        else{
            cout<<(c=query(rt,1,mx,x,y))<<'\n';
        }
    }
}
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