# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
348300 | kasparovian | 원숭이와 사과 나무 (IZhO12_apple) | C++14 | 630 ms | 232940 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// '-.-'
// () __.'.__
// .-:--:-. |_______|
// () \____/ \=====/
// /\ {====} )___(
// (\=, //\\ )__( /_____\
// __ |'-'-'| // .\ ( ) /____\ | |
// / \ |_____| (( \_ \ )__( | | | |
// \__/ |===| )) `\_) /____\ | | | |
// /____\ | | (/ \ | | | | | |
// | | | | | _.-'| | | | | | |
// |__| )___( )___( /____\ /____\ /_____\
// (====) (=====) (=====) (======) (======) (=======)
// }===={ }====={ }====={ }======{ }======{ }======={
// (______)(_______)(_______)(________)(________)(_________)
//
// Credits :- Joan G. Stark
//|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|
//| AUTHOR - KASPAROVIAN |
//|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define frr(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define fr first
#define sc second
#define mk make_pair
#define endl '\n'
#define MOD 1000000007
#define in insert
#define sz(x) (ll)(x).size()
#define mem(a,b) memset(a,b,sizeof(a))
// #define int long long
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef long double ld;
typedef pair<ll,ll> pl;
typedef pair<int,int> pi;
typedef pair<int,pi> ppi;
typedef vector<vi> graph;
template<class T> void mxi(T & a, const T & b) { a = max(a, b); }
template<class T> void mni(T & a, const T & b) { a = min(a, b); }
ld EPS=1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define SHUF(v) shuffle(all(v), RNG);
// Use mt19937_64 for 64 bit random numbers.
const int SZ=1e9+100;
struct node
{
int cnt,lazy; node *c[2];
node()
{
cnt=lazy=0;
c[0]=c[1]=NULL;
}
#define mid ((l+r)>>1)
void push(int l,int r)
{
if(lazy)
{
cnt=r-l+1;
if(l!=r)
{
frr(i,2)
{
if(!c[i])c[i]=new node();
c[i]->lazy=1;
}
c[0]->cnt=mid-l+1;
c[1]->cnt=r-mid;
}
}
lazy=0;
}
void upd(int lq,int rq,int l=0,int r=SZ-1)
{
push(l,r);
if(lq<=l&&rq>=r)
{
lazy=1; push(l,r); return;
}
if(rq<=mid)
{
if(!c[0])c[0]=new node();
c[0]->upd(lq,rq,l,mid);
}
else if(lq>=mid+1)
{
if(!c[1])c[1]=new node();
c[1]->upd(lq,rq,mid+1,r);
}
else
{
frr(i,2)
{
if(!c[i])c[i]=new node();
}
c[0]->upd(lq,rq,l,mid);
c[1]->upd(lq,rq,mid+1,r);
}
cnt=0;
frr(i,2)if(c[i])cnt+=c[i]->cnt;
}
int qu(int lq,int rq,int l=0,int r=SZ-1)
{
push(l,r);
if(rq<l||r<lq)return 0;
else if(lq<=l&&rq>=r)
{
return cnt;
}
int ans=0;
if(c[0])ans+=c[0]->qu(lq,rq,l,mid);
if(c[1])ans+=c[1]->qu(lq,rq,mid+1,r);
return ans;
}
};
void solve()
{
int q,c=0; cin>>q;
node Seg;
while(q--)
{
int d,x,y; cin>>d>>x>>y;
if(d==1)
{
int ans=Seg.qu(x+c,y+c);
c=ans;
cout<<ans<<endl;
}
else
{
Seg.upd(x+c,y+c);
}
}
}
signed main()
{
fast;
int t,tab;
t=1;
tab=t;
while(t--)
{
//cout<<"Case #"<<(tab-t)<<": ";
solve();
}
cerr<<runtime();
}
//APPROACHING A QUESTION
//+ Think of binary search (max of min etc also if n<=2*10^5)
//+ Think of common dp states (Even if it appears as maths but constraints are small)
//+ Check constraints
//+ Keep calm and enjoy the question
//+ Be sure to remove MOD from binpow (if needed)
//+ Try bidirectional analysis for constructive questions
//+ If given some sequence try thinking of prefix sums
//+ If constraints are too large maybe its simple maths
//+ In questions with binary operations think of bits independently and also the change pattern
//+ If two or more binary operations are given mostly there is a relation between them and an arithmatic operator
컴파일 시 표준 에러 (stderr) 메시지
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