sort and show the number of digits - c++

Question

I want to my program can sort the inputted integer and compute the number of any integer that inputted and I don't know where should write the cout of c
example
a[9]={2,3,2,6,6,3,5,2,2}
the number of 2 is 4
the number of 3 is 2
the number of 6 is 2
.
.
please fix this code
int main()
{
cout << "please enter the number of digites :" << endl;
int n;
cin>>n;
int a[n];
cout<<"enter numbers :"<<endl;
for(int i=0;i<n;i++)
cin>>a[i];
int i,j;
for(i=0;i<n-1;i++)
{
for(j=0;j<n-i-1;j++)
if(a[j]>a[j+1])
{
int temp;
temp=a[j+1];
a[j+1]=a[j];
a[j]=temp;
}
}
int c;
for(int m=0;m<n;m++)
{
if(a[m]==a[m+1])
c++;
else
c=0;
}
return 0;
}

Answer 1

Read through my solution, I've commented the parts I've changed. I tidied it up a little.
To answer your question: you should print the output (frequency of an integer in array) before you reset the count variable to 1. This will work because we have sorted the array, and will not have to look ahead for more occurrences of the current number.
[EDIT] I also added this above your code:
#include <iostream>
#include <vector>
using namspace std;
Full Solution
#include <iostream>
#include <vector>
using namespace std;
int main() {
// Get input
int n;
cout << "Please enter the number of digits: ";
cin>>n;
vector<int> a;
cout << "Enter " << n << " numbers: " << endl;
for(int i=0;i<n;i++) {
int temp;
cin >> temp;
a.push_back(temp);
}
// Sort input
int i,j;
for (i = 0; i < a.size(); i++) {
for(j = 0; j < a.size()-i-1; j++) {
if(a[j] > a[j+1]) {
int temp;
temp=a[j+1];
a[j+1]=a[j];
a[j]=temp;
}
}
}
// If an element is in an array
// we can not have 0 occurrences
// of that element, hence count
// must start at 1
int count = 1;
// Int to count
int current = a[0];
// Ouput if we have reset the count,
// or if it is the last iteration
bool output;
// Loop through array
for (int i = 1; i < a.size(); i++) {
output = false; // Reset output if we have printed
if (a[i] == current) {
// If current int and the element next to it are the same,
// increase the count
count++;
} else {
// If current and next are different,
// we need to show the frequency,
// and then reset count to 1
cout << current << " occurs " << count << " times" << endl;
count = 1;
current = a[i];
}
}
// Output one last time, for last int in sorted set
cout << current << " occurs " << count << " times" << endl;
return 0;
}
If this doesn't help, go and read this page, it is a solution in C, but can be adapted to C++ easily. https://codeforwin.org/2015/07/c-program-to-find-frequency-of-each-element-in-array.html This will help you understand and write the task. They take you step-by-step through the algorithm.

Answer 2

This is a typical use-case for a std::map. A std::map<char,int> lets you easily count the frequency of charaters (its easier to treat the user input as characters instead of converting it to numbers).
This is basically all you need:
#include <iostream>
#include <iterator>
#include <map>
int main(){
std::istream_iterator<char> it( std::cin );
std::istream_iterator<char> end_of_input;
std::map<char,int> data;
while (it != end_of_input ) data[*(it++)]++;
for (const auto& e : data) std::cout << e.first << " " << e.second << "\n";
}
This is probably a lot at once, so lets go one by one.
std::istream_iterator<char> lets you extract characters from a stream as if you are iterating a container. So the while iteratates std::cin until it reaches the end of the input. Then *(it++) increments the iterator and returns the character extracted from the stream. data[x]++ accesses the value in the map for the key x and increments its value. If there is no value in the map yet for the key, it gets default initialized to 0.
For input: 11223 it prints
1 2
2 2
3 1
Your code has some issues, not sure if I can catch them all...
You are using VLA (variable lenght arrays) here: int a[n];. This is a compiler extension and not standard c++.
You access the array out of bounds. When i == 0 then j goes up to j<n-i-1 == n-1 and then you access a[j+1] == a[n], but the last valid index into the array is n-1. Same problem in the other loop (a[m+1]).
Assuming your sorting works, the last loop almost gives you the number of elements, but not quite, to fix it you can change it to...
int current = a[0];
int counter = 1;
for(int m=1;m<n;m++) {
if(a[m] == current) {
counter++;
} else {
std::cout << current << " appears " << counter << " times" << endl;
counter=1; // note: minimum freq is 1 not 0
current = a[m];
}
}