Submission #443270

#	Time	Username	Problem	Language	Result	Execution time	Memory
443270		minhcool	Examination (JOI19_examination)	C++17	0 / 100	1 ms	204 KiB

examination

#include<bits/stdc++.h>
using namespace std;

#define uint unsigned int
#define int long long
#define fi first
#define se second
#define pb push_back
#define hash hash1
#define index indexx
#define foru(i, l, r) for(int i = l; i <= r; i++)
#define ford(i, r, l) for(int i = r; i >= l; i--)

typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;

const int N = 2e5 + 5;

const int oo = 1e18 + 7, mod = 1e9 + 7;

const int hashy = 1e9 + 123;

/*
I think this is one of the easiest problems that you can give a guy if he knows segtree 2D
*/

// I think this will run as slow as f*ck, but it is sure O(1) lol

struct hash{
	int operator()(const ii&a)const{
		return (a.fi << 30) | a.se;
	}
};

int n, a[N];

int cnt;
uint IT[N * 30 * 30];
unordered_map<ii, int, hash> index;

int cal(int x, int y){
	if(index.find({x, y}) != index.end()) return index[{x, y}];
	cnt++;
	return index[{x, y}] = cnt;
}

void updy(int ind, int lx, int rx, int ly, int ry, int posy, int val){
	if(ly == ry){
		IT[ind] += val;
		return;
	}
	int mid = (ly + ry) >> 1;
	int temp1 = cal((lx << 30LL) + rx, (ly << 30LL) + mid), temp2 = cal((lx << 30LL) + rx, ((mid + 1) << 30LL) + ry);
	if(posy <= mid) updy(temp1, lx, rx, ly, mid, posy, val);
	else updy(temp2, lx, rx, mid + 1, ry, posy, val);
	IT[ind] = IT[temp1] + IT[temp2];
}

void updx(int id, int lx, int rx, int posx, int val){
	//int tmp = cal(lx, rx, 1, (1LL << 30), posx, po)
	//updy()
}

void process(){
	
}

signed main(){
	ios_base::sync_with_stdio(0);
	process();
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 20
• Subtask 3: 0 / 21
• Subtask 4: 0 / 57