Submission #7256

#	Time	Username	Problem	Language	Result	Execution time	Memory
7256		Rhak	Game (IOI13_game)	C++98	80 / 100	13000 ms	247000 KiB

game

#include "game.h"
#include<stdio.h>

long long gcd2(long long x, long long y){ return y==0? x: gcd2(y, x%y); }

struct _node{
	_node *d[4];
	long long GCD;
	_node(long long GCD):GCD(GCD){ d[0] = d[1] = d[2] = d[3] = 0; }
};

struct _IndexTree{
	int S, E;
	_node *root;
	_IndexTree(int S,int E):S(S), E(E){ root = new _node(0);}
	void update(_node* n, int s, int e, int p, long long N, _IndexTree** arr)
	{
		if( p < s || e < p) return;
		else if( p == s && e == p){
			if( N != -1) n->GCD = N;
			else{
				long long fn = 0;
				for(int i = 0; i < 4; i++){
					if(arr[i]) fn = gcd2(fn, arr[i]->read(s,e));
				}
				n->GCD = fn;
			}
			return;
		}
		long long t[4] = {0}, fn = 0; 
		for(int i = 0; i < 4; i++) t[i] = n->d[i]? n->d[i]->GCD:0;
		if( e >= s+4){
			int mr = ( (long long)s*3 + e) >> 2;
			if( s <= p && p <= mr ){
				if( !n->d[0] ) n->d[0] = new _node(0);
				update(n->d[0], s, mr, p, N, arr);
				t[0] = n->d[0]->GCD;
			}
			for(long long i = 1; i < 4; i++){
				int ml = (( s*(4-i) + e*i) >> 2) + 1;
				int mr = ( s*(3-i) + e*(i+1)) >> 2;
				if( ml <= p && p <= mr ){
					if( !n->d[i] ) n->d[i] = new _node(0);
					update(n->d[i], ml, mr, p, N, arr);
					t[i] = n->d[i]->GCD;
				}
			}
		}
		else{
			for(int j = s, i = 0; j <= e; i++, j++){
				if( j != p) continue;
				if( !n->d[i] ) n->d[i] = new _node(0);
				update(n->d[i], j, j, p, N, arr);
				t[i] = n->d[i]->GCD;
			}
		}		
		for(int i = 0; i < 4; i ++) fn = gcd2(fn, t[i]);
		n->GCD = fn;
	}
	void update(int p, long long N)
	{
		_IndexTree* arr[4] = {0};
		update(root, S, E, p, N, arr);
	}

	void update(int p, _IndexTree** arr)
	{
		update(root, S, E, p, -1, arr);
	}

	long long read(_node *n, int s, int e, int p, int q)
	{
		if( !n ) return 0;
		if( q < s || e < p) return 0;
		else if( p <= s && e <= q){
			return n->GCD;
		}
		int m = (s+e) >> 1;
		long long t , fn = 0;
		if( e >= s+4){
			int mr = ( (long long)s*3 + e) >> 2;
			t = n->d[0]? read(n->d[0], s, mr, p, q):0;
			fn = gcd2(fn, t);
			for(long long i = 1; i < 4; i++){
				int ml = (( s*(4-i) + e*i) >> 2) + 1;
				int mr = ( s*(3-i) + e*(i+1)) >> 2;
				t = n->d[i]? read(n->d[i], ml, mr, p, q):0;
				fn = gcd2(fn, t);
			}
		}
		else{
			for(int j = s, i = 0; j <= e; i++, j++){
				t = n->d[i]? read(n->d[i], j, j, p, q):0;
				fn = gcd2(fn, t);
			}
		}		
		return fn;
	}

	long long read(int s,int e){
		return read(root, S, E, s, e);
	}
};

struct node{
	node *d[4];
	_IndexTree *IT;
	node(int C){
		d[0] = d[1] = d[2] = d[3] = 0;
		IT = new _IndexTree(0, C);
	}
};

struct IndexTree{
	int S, E, C;
	node* root;
	IndexTree(int S, int E, int C):S(S), E(E), C(C){
		root = new node(C);
	}
	
	void update(node* n, int s, int e, int p, int q, long long N)
	{
		if( p < s || e < p) return;
		if( p == s && e == p){
			n->IT->update(q, N);
			return;
		}
		long long t[4] = {0}, fn = 0; 
		if( e >= s+4){
			int mr = ( (long long)s*3 + e) >> 2;
			if( s <= p && p <= mr ){
				if( !n->d[0] ) n->d[0] = new node(C);
				update(n->d[0], s, mr, p, q, N);
			}
			for(long long i = 1; i < 4; i++){
				int ml = (( s*(4-i) + e*i) >> 2) + 1;
				int mr = ( s*(3-i) + e*(i+1)) >> 2;
				if( ml <= p && p <= mr ){
					if( !n->d[i] ) n->d[i] = new node(C);
					update(n->d[i], ml, mr, p, q, N);
				}
			}
		}
		else{
			for(int j = s, i = 0; j <= e; i++, j++){
				if( j != p) continue;
				if( !n->d[i] ) n->d[i] = new node(C);
				update(n->d[i], j, j, p, q, N);
			}
		}
		_IndexTree *arr[4] = {0};
		for(int i = 0; i < 4; i++) arr[i] = n->d[i]? n->d[i]->IT: 0;
		n->IT->update(q, arr);
	}
	void update(int p, int q, long long N)
	{
		update(root, S, E, p, q, N);
	}

	long long read(node *n, int s, int e, int p, int q, int u, int v)
	{
		if( !n ) return 0;
		if( q < s || e < p) return 0;
		else if( p <= s && e <= q){
			return n->IT->read(u, v);
		}
		long long t, fn = 0;
		if( e >= s+4){
			int mr = ((long long)s*3 + e) >> 2;
			fn = gcd2(fn, n->d[0]? read(n->d[0], s, mr, p, q, u, v):0);
			for(long long i = 1; i < 4; i++){
				int ml = (( s*(4-i) + e*i) >> 2) + 1;
				int mr = ( s*(3-i) + e*(i+1)) >> 2;
				fn = gcd2(fn, n->d[i]? read(n->d[i], ml, mr, p, q, u, v):0);
			}
		}
		else{
			for(int j = s, i = 0; j <= e; i++, j++){
				t = n->d[i]? read(n->d[i], j, j, p, q, u, v):0;
				fn = gcd2(fn, t);
			}
		}
		return fn;
	}
	long long read(int P,int Q,int U, int V)
	{
		return read(root, S, E, P, Q, U, V);
	}
}*IT;

void init(int R, int C){
	IT = new IndexTree(0, R-1, C-1);
}

void update(int R, int C, long long K){
	IT->update(R, C, K);
}

long long calculate(int P, int Q, int U, int V){
	return IT->read(P, U, Q, V);
}

• Subtask 1: 10 / 10
• Subtask 2: 27 / 27
• Subtask 3: 26 / 26
• Subtask 4: 17 / 17
• Subtask 5: 0 / 20