Algorithm
Problem Name: beecrowd | 2542
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/2542
Iu-Di-Oh!
By Ricardo Oliveira, UFPR 
Brazil
Timelimit: 1
Iu-di-oh! is a card game really popular among kids! Every Iu-di-oh! player has his own deck containing many cards. Each card contains N attributes (such as power, speed, smartness, etc.). Attributes are numbered from 1 to N and are given as positive integers.
A match of Iu-di-oh! is always played by two players. At the beginning of the match, each player chooses exactly one card from his deck. Then, an attribute is randomly chosen. The player whose the chosen attribute is greater in the card he choose wins the match. If the such attribute is equal in both cards, there is a tie.
Marcos and Leonardo are in the big final of the Brazilian Iu-di-oh! championship. The great prize is a Dainavision (that is almost as good as a Plaisteition 2!). Given the deck of both players, the card each one chooses and the chosen attribute, determine the winner!
Input
The input contains several test cases. The first line of each test case contains an integer N (1 ≤ N ≤ 100), the number of attributes each card contains. The second line contains two integers M and L (1 ≤ M, L ≤ 100), the number of cards in Marcos’ and Leonardo’s deck, respectively.
Next M lines describe Marcos’ deck. His cards are numbered from 1 to M, and i-th line describes the i-th card. Each line contains N integers ai,1,ai,2,..., ai,N (1 ≤ ai,j ≤ 109). Integer ai,j indicates the j-th attribute of the i-th card.
Next L lines describe Leonardo’s deck. His cards are numbered from 1 to L and are described in the same way as Marcos’ deck.
Next line contains two integers CM and CL (1 ≤ CM ≤ M, 1 ≤ CL ≤ L), the cards chosen by Marcos and Leonardo, respectively. Finally, the last line contains an integer A (1 ≤ A ≤ N) indicating the chosen attribute.
The input ends with end-of-file (EOF).
Output
For each test case, print a line containing “Marcos” if Marcos wins the match, “Leonardo” if Leonardo wins the match, or “Empate” in the case of a tie (without quotes).
| Input Sample | Output Sample |
|
3 |
Marcos |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main(void)
{
int n, m, l, i, j, c1, c2, at, x, y;
while (scanf("%i %i %i", &n, &m, &l) != EOF)
{
int am[m][n], al[l][n];
for (i = 0; i < m; ++i)
{
for (j = 0; j < n; ++j)
scanf("%i", &am[i][j]);
}
for (i = 0; i < l; ++i)
{
for (j = 0; j < n; ++j)
scanf("%i", &al[i][j]);
}
scanf("%i %i %i", &c1, &c2, &at);
x=am[c1-1][at-1];
y=al[c2-1][at-1];
if (x > y)
printf("Marcos\n");
else if (y > x)
printf("Leonardo\n");
else
printf("Empate\n");
}
return 0;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include<iostream>
#include<vector>
using namespace std;
int main()
{
int n;
while(cin >> n)
{
int m,l;
cin >> m >> l;
vector < int> M[m];
vector <int> L[l];
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
int a;
cin >> a;
M[i].push_back(a);
}
}
for(int i = 0; i < l; i++)
{
for(int j = 0; j < n; j++)
{
int a;
cin >> a;
L[i].push_back(a);
}
}
int MM,LL;cin >> MM >> LL;
int A;
cin >> A;
long long s = M[MM-1][A-1];
long long S = L[LL-1][A-1];
if(s>S) cout << "Marcos\n";
else if(s < S> cout << "Leonardo\n";
else cout << "Empate\n";
}
return 0;
}
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n = sc.nextInt(); //numero de atributos de cada carta
int m = sc.nextInt(); //numero de cartas de Marcos
int l = sc.nextInt(); //numero de cartas de Leonardo
long[][] cartasM = new long[m][n]; //cartas de Marcos
//entra cartas de Marcos
for(int i = 0 ; i < m ; i++)
for(int j = 0 ; j < n ; j++)
cartasM[i][j] = sc.nextLong();
long[][] cartasL = new long[l][n]; //cartas de Leonardo
//entra cartas de Leonardo
for(int i = 0 ; i < m ; i++)
for(int j = 0 ; j < n ; j++)
cartasL[i][j] = sc.nextLong();
int cm; //cartas escolhidas por Marcos
cm = sc.nextInt()-1;
int cl; //cartas escolidas por Leonardo
cl = sc.nextInt()-1;
int a; //atributo sorteado
a = sc.nextInt()-1;
long Marcos = cartasM[cm][a];
long Leonardo = cartasL[cl][a];
if (Marcos > Leonardo)
System.out.println("Marcos");
else if (Leonardo > Marcos)
System.out.println("Leonardo");
else
System.out.println("Empate");
}
sc.close();
}
}
Input
#4 Code Example with Javascript Programming
Code -
Javascript Programming
var input = require('fs').readFileSync('/dev/stdin', 'utf8');
var lines = input.split('\n');
var prompt = function(texto) { return lines.shift();};
while (true) {
var attck = parseInt(prompt());
if (isNaN(attck)) {
break;
}
var [cmarcos, cleo] = prompt().split(" ").map(Number);
var bmarcos = [];
var bleo = [];
if (attck == 1) {
for (let i = 0; i < cmarcos; i++) {
let card = parseInt(prompt());
bmarcos.push(card);
}
for (let i = 0; i < cleo; i++) {
let cartas = parseInt(prompt());
bleo.push(cartas);
}
} else {
for (let i = 0; i < cmarcos; i++) {
let cartas = prompt().split(" ").map(Number);
bmarcos.push(cartas);
}
for (let i = 0; i < cleo; i++) {
let cartas = prompt().split(" ").map(Number);
bleo.push(cartas);
}
}
var [marcosescolha, leoescolha] = prompt().split(" ").map(Number);
var attr = parseInt(prompt());
if (bmarcos[marcosescolha - 1][attr - 1] > bleo[leoescolha - 1][attr - 1]) {
console.log("Marcos");
} else if (
bmarcos[marcosescolha - 1][attr - 1] < bleo[leoescolha - 1][attr - 1]
) {
console.log("Leonardo");
} else {
console.log("Empate");
}
}
Input
#5 Code Example with Python Programming
Code -
Python Programming
while True:
try:
input()
e = str(input()).split()
ncm = int(e[0])
ncl = int(e[1])
cm = []
cl = []
for i in range(ncm): cm.append([int(x) for x in str(input()).split()])
for i in range(ncl): cl.append([int(x) for x in str(input()).split()])
e = str(input()).split()
ms = int(e[0])
lu = int(e[1])
a = int(input())
if cm[ms-1][a-1] > cl[lu-1][a-1]: print('Marcos')
elif cm[ms-1][a-1] < cl[lu-1][a-1]: print('Leonardo')
else: print('Empate')
except EOFError: break
Input