Submission #448120

#	Time	Username	Problem	Language	Result	Execution time	Memory
448120		ignaciocanta	Sateliti (COCI20_satellti)	C++14	110 / 110	542 ms	34920 KiB

Main

#include <bits/stdc++.h>
 
using namespace std;
 
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
//template<class T> using Tree = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
 
using tint = long long;
using ld = long double;
 
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)	
 
using pi = pair<int,int>;
using pl = pair<tint,tint>;
using vi = vector<int>;
using vl = vector<tint>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using vb = vector<bool>;
 
#define pb push_back
#define pf push_front
#define rsz resize
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sz(x) (int)(x).size()
#define ins insert
 
#define f first
#define s second
#define mp make_pair
 
#define DBG(x) cerr << #x << " = " << x << endl;
 
const int MOD = 1e9+7; 
const int mod = 998244353;
const int MX = 1005;
const tint INF = 1e18; 
const int inf = 1e9;
const ld PI = acos(ld(-1)); 
const ld eps = 1e-5;
 
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
 
template<class T> void remDup(vector<T> &v){ 
	sort(all(v)); v.erase(unique(all(v)),end(v));
}
 
template<class T> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; 
} 
template<class T> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; 
}
 
bool valid(int x, int y, int n, int m){
	return (0<=x && x<n && 0<=y && y<m);
}
 
int cdiv(int a, int b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba
int fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redondea p abajo
 
void NACHO(string name = ""){
	ios_base::sync_with_stdio(0); cin.tie(0);
	if(sz(name)){
		freopen((name+".in").c_str(), "r", stdin);
		freopen((name+".out").c_str(), "w", stdout);
	}
}

struct mi {
 	int v; explicit operator int() const { return v; } 
	mi() { v = 0; }
	mi(tint _v):v(_v%MOD) { v += (v<0)*MOD; }
};
mi& operator+=(mi& a, mi b) { 
	if ((a.v += b.v) >= MOD) a.v -= MOD; 
	return a; }
mi& operator-=(mi& a, mi b) { 
	if ((a.v -= b.v) < 0) a.v += MOD; 
	return a; }
mi operator+(mi a, mi b) { return a += b; }
mi operator-(mi a, mi b) { return a -= b; }
mi operator*(mi a, mi b) { return mi((tint)a.v*b.v); }
mi& operator*=(mi& a, mi b) { return a = a*b; }
mi pow(mi a, tint p) { assert(p >= 0); 
	return p==0?1:pow(a*a,p/2)*(p&1?a:1); }
mi inv(mi a) { assert(a.v != 0); return pow(a,MOD-2); }
mi operator/(mi a, mi b) { return a*inv(b); }

char a[MX][MX];
mi hsh[2*MX][2*MX];
mi sh[2*MX][2*MX];
mi pw1[2*MX], pw2[2*MX];

int P1 = 9973, P2 = 10007;

mi query(int a, int b, int c, int d){
	return sh[c][d] - sh[a-1][d] - sh[c][b-1] + sh[a-1][b-1];
}

int main(){
	NACHO();
	// trabajar con los shifts es molesto. Por eso creamos una nueva matriz de tamaño 2*n y 2*m
	// para que un shift corresponda a cualquier submatriz de n*m en esta nueva matriz.
	// como comparamos lexicograficamente dos submatrices?
	// podemos hacerlo con hashing en O(logn).
	int n, m; cin >> n >> m;
	pw1[0] = pw2[0] = 1;
	FOR(i, 1, 2*n+1) pw1[i] = pw1[i-1] * P1;
	FOR(i, 1, 2*m+1) pw2[i] = pw2[i-1] * P2;
	F0R(i, n) cin >> a[i];
	F0R(i, 2*n){
		F0R(j, 2*m){
			hsh[i][j] = a[i%n][j%m] * pw1[i+1] * pw2[j+1];
		}
	}
	FOR(i, 1, 2*n+1){
		FOR(j, 1, 2*m+1){
			sh[i][j] = sh[i-1][j] + sh[i][j-1] - sh[i-1][j-1] + hsh[i-1][j-1];
		}
	}
	int posX = 1, posY = 1;
	FOR(i, 1, n+2){
		FOR(j, 1, m+2){
			int low = -1, high = n;
			while(high-low > 1){
				int mid = low+(high-low)/2;
				mi curHshErase = query(i, j, i+mid, j+m-1) * (pw1[posX] * pw2[posY]);
				mi bestHshErase = query(posX, posY, posX+mid, posY+m-1) * (pw1[i] * pw2[j]);
				if((int)curHshErase == (int)bestHshErase) low = mid;
				else high = mid;
			}
			int X = high;
			low = -1, high = m;
			while(high-low > 1){
				int mid = low+(high-low)/2;
				mi curHshErase = query(i, j, i+X, j+mid) * (pw1[posX] * pw2[posY]);
				mi bestHshErase = query(posX, posY, posX+X, posY+mid) * (pw1[i] * pw2[j]);
				if((int)curHshErase == (int)bestHshErase) low = mid;
				else high = mid;
			}
			int Y = high;
			if(a[(i+X-1)%n][(j+Y-1)%m] < a[(posX+X-1)%n][(posY+Y-1)%m]){
				posX = i, posY = j;
			} 
		}
	}
	FOR(i, posX-1, posX+n-1){
		FOR(j, posY-1, posY+m-1){
			cout << a[i%n][j%m];
		}
		cout << "\n";
	}
}
/*
3 5
..*.*
**..*
****.
*/

Compilation message (stderr)

Main.cpp: In function 'void NACHO(std::string)':
Main.cpp:74:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   74 |   freopen((name+".in").c_str(), "r", stdin);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Main.cpp:75:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   75 |   freopen((name+".out").c_str(), "w", stdout);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 10 / 10
• Subtask 2: 40 / 40
• Subtask 3: 60 / 60