Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1171
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1171
Number Frequence
Adapted by Neilor Tonin, URI 
Brazil
Timelimit: 1
In this problem your job is to read some positive and integer numbers and print how many times each number appears in the input, you must write each of the distinct values that appear in the input, ordering by ascending value.
Input
The input contains only one test case. The first line of input contains one integer N, which indicates the quantity of numbers that will be read to X (1 ≤ X ≤ 2000) in the sequence. Each number don't appears more than 20 times in the problem input.
Output
Print the output according to the example provided below, indicating how many times each number appears in the input file, by ascending order of value.
| Input Sample | Output Sample |
|
7 |
4 aparece 1 vez(es) |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main(void) {
int q, i, n;
int lista[2000];
for (i = 0; i < 2000; ++i)
lista[i] = 0;
scanf("%d", &q);
for (i = 0; i < q; ++i) {
scanf("%d", &n);
++lista[n-1];
}
for (i = 0; i < 2000; ++i)
if (lista[i] != 0)
printf("%d aparece %d vez(es)\n", i+1, lista[i]);
return 0;
}
Input
8
10
8
260
4
10
10
Output
8 aparece 2 vez(es)
10 aparece 3 vez(es)
260 aparece 1 vez(es)
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
int main(void) {
int q, i, n;
int lista[2000];
for (i = 0; i < 2000; ++i)
lista[i] = 0;
cin >> q;
for (i = 0; i < q; ++i) {
cin >> n;
++lista[n-1];
}
for (i = 0; i < 2000; ++i)
if (lista[i] != 0)
cout << i+1 << " aparece " << lista[i] << " vez(es)" << endl;
return 0;
}
Input
8
10
8
260
4
10
10
Output
8 aparece 2 vez(es)
10 aparece 3 vez(es)
260 aparece 1 vez(es)
#3 Code Example with Java Programming
Code -
Java Programming
import java.io.IOException;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
while (sc.hasNextInt()) {
int n = 0, num = 0, cont = 0, aux = 0, cont2 = 0;
n = sc.nextInt();
int[] vet = new int[n];
for (int i = 0; i < n; i++) {
vet[i] = sc.nextInt();
}
for (int i = 0; i < vet.length; i++) {
for (int j = 0; j < vet.length - 1; j++) {
if (vet[j] > vet[j + 1]) {
aux = vet[j];
vet[j] = vet[j + 1];
vet[j + 1] = aux;
}
}
}
for (int i = 0; i < vet.length; i++) {
cont = 0;
num = vet[i];
for (int j = 0; j < vet.length; j++) {
if (vet[i] == vet[j] && vet[i] != 0 && vet[j] != 0 && vet[j] != cont2) {
cont++;
}
}
cont2 = vet[i];
if (cont != 0 && vet[i] != 0) {
System.out.println(num + " aparece " + cont + " vez(es)");
}
}
}
}
}
Input
8
10
8
260
4
10
10
Output
8 aparece 2 vez(es)
10 aparece 3 vez(es)
260 aparece 1 vez(es)
#4 Code Example with Python Programming
Code -
Python Programming
n = int(input())
v = []
for i in range(n):
v.append(int(input()))
v.sort()
for i, j in enumerate(v):
if i == 0:
print('{} aparece {} vez(es)'.format(j, v.count(j)))
elif j != u: print('{} aparece {} vez(es)'.format(j, v.count(j)))
u = j
Input
8
10
8
260
4
10
10
Output
8 aparece 2 vez(es)
10 aparece 3 vez(es)
260 aparece 1 vez(es)