Algorithm
Problem Name: beecrowd | 1435
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1435
Square Matrix I
BrazilTimelimit: 2
Write a program that read an integer number N (0 ≤ N ≤ 100) that correspond to the order of a Bidimentional array of integers, and build the Array according to the above example.
Input
The input consists of several integers numbers, one per line, corresponding to orders from arrays to be built. The end of input is indicated by zero (0).
Output
For each integer number of input, print the corresponding array according to the example. (the values of the arrays must be formatted in a field of size 3 right justified and separated by a space. None space must be printed after the last character of each row of the array. A blank line must be printed after each array.
| Sample Input | Sample Output |
|
1 |
1 |
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
var input = require('fs').readFileSync('/dev/stdin', 'utf8');
var lines = input.split('\n');
let contador = parseInt(lines.shift());
let vetor = [];
let matriz = [];
while (contador !== 0) {
matriz = new Array(contador);
for (let i = 0; i < contador / 2; i++) {
vetor = new Array(contador);
for (let j = 0; j < contador / 2; j++) {
const value = Math.min(i, j) + 1;
vetor[j] = value < 10 ? ` ${value}` : `${value}`;
vetor[contador - 1 - j] = vetor[j];
}
matriz[i] = ` ${vetor.join(' ')}`;
matriz[contador - 1 - i] = matriz[i];
}
console.log(`${matriz.join('\n')}`);
contador = parseInt(lines.shift());
console.log('');
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
int main(){
int n, i, j, mat[100][100] = {}, temp = 1;
bool is = false;
while(cin >> n and n != 0){
for (i = 0;i < n;i++) {
for (j = 0;j < n;j++) {
mat[i][j] = 1;
if (i > 0 and j > 0 and i < n-1 and j < n-1) {
mat[i][j] += mat[temp][temp];
}
if (j == i) {
is = true;
}
}
if(is == true){
temp+=1;
is = false;
}
}
for (i = 0;i < n;i++) {
for (j = 0;j < n;j++> {
cout << mat[i][j] << " ";
}
cout << endl;
}
}
return 0;
}