Algorithm
Problem Name: beecrowd | 1179
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1179
Array Fill IV
Adapted by Neilor Tonin, URI 
Brazil
Timelimit: 1
In this problem you need to read 15 numbers and must put them into two different arrays: par if the number is even or impar if this number is odd. But the size of each of the two arrrays is only 5 positions. So every time you fill one of two arrays, you must print the entire array to be able to use it again for the next numbers that are read. At the end, all remaining numbers of each one of these two arrays must be printed beggining with the odd array. Each array can be filled how many times are necessary.
Input
The input contains 15 integer numbers.
Output
Print the output like the following example.
| Input Sample | Output Sample |
|
1 |
par[0] = 4 |
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int a1[5],a2[5];
int count1 = 0,count2 = 0,num;
for(int i = 0; i < 15; i++)
{
cin >> num;
if(num%2 == 0)
{
a1[count1] = num;
count1++;
}
if(num%2!=0)
{
a2[count2]=num;
count2++;
}
if(count2==5||i==14)
{
for(int o = 0; o < count2; o++)
{
printf("impar[%d] = %d\n",o,a2[o]);
}
count2=0;
}
if(count1==5||i==14)
{
for(int i = 0; i < count1; i++)
{
printf("par[%d] = %d\n",i,a1[i]>;
}
count1=0;
}
}
return 0;
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const input = require('fs').readFileSync('/dev/stdin', 'utf8');
const lines = input.split('\n');
function clear(n){
while(n.length){
n.pop();
}
return n;
}
let par = []; let impar = [];
for(var i = 0; i < 15; i++){
let n = parseInt(lines.shift());
if(n % 2 !== 0){
impar.push(n);
if(impar.length === 5){
for(var m = 0; m < 5; m++){
console.log(`impar[${m}] = ${impar[m]}`);
}
clear(impar);
}
}
else{
par.push(n);
if(par.length === 5){
for(var m = 0; m < 5; m++){
console.log(`par[${m}] = ${par[m]}`);
}
clear(par);
}
}
}
if(impar.length !== 0){
for(var i = 0; i < impar.length; i++){
console.log(`impar[${i}] = ${impar[i]}`);
}
}
if(par.length !== 0){
for(var i = 0; i < par.length; i++){
console.log(`par[${i}] = ${par[i]}`>;
}
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
par = []; impar = []
for i in range(0, 15):
n = int(input())
if n % 2 != 0:
impar += [n]
if len(impar) == 5:
for m in range(0, 5):
print(f'impar[{m}] = {impar[m]}')
impar.clear()
else:
par += [n]
if len(par) == 5:
for m in range(0, 5):
print(f'par[{m}] = {par[m]}')
par.clear()
if len(impar) != 0:
for i in range(0, len(impar)):
print(f'impar[{i}] = {impar[i]}')
if len(par) != 0:
for i in range(0, len(par)):
print(f'par[{i}] = {par[i]}')
Input
Output