Problem:
Counting Sort

Counting sort can be used for sorting elements in an array which each of the n input elements is an integer in the range 0 to k. The idea of counting sort is to determine, for each input element x, the number of elements less than x as C[x]. This information can be used to place element x directly into its position in the output array B. This scheme must be modified to handle the situation in which several elements have the same value. Please see the following pseudocode for the detail:

Counting-Sort(A, B, k)
1 for i = 0 to k
2 do C[i] = 0
3 for j = 1 to length[A]
4 do C[A[j]] = C[A[j]]+1
5 /* C[i] now contains the number of elements equal to i */
6 for i = 1 to k
7 do C[i] = C[i] + C[i-1]
8 /* C[i] now contains the number of elements less than or equal to i */
9 for j = length[A] downto 1
10 do B[C[A[j]]] = A[j]
11 C[A[j]] = C[A[j]]-1

Write a program which sorts elements of given array ascending order based on the counting sort.

Input

The first line of the input includes an integer n, the number of elements in the sequence.

In the second line, n elements of the sequence are given separated by spaces characters.

Output

Print the sorted sequence. Two contiguous elements of the sequence should be separated by a space character.

Constraints

1 ≤ n ≤ 2,000,000

0 ≤ A[i] ≤ 10,000

Sample Input 1

7
2 5 1 3 2 3 0

Sample Output 1

0 1 2 2 3 3 5

Solution in C++11 (judge verdict: Accepted)

#include <stdio.h>
#include <sstream>
#include <string.h>
#include <vector>
#include <map>
#include <algorithm>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <limits>
#include <iomanip>
#include <ctype.h>
#define _USE_MATH_DEFINES
#include <iostream>
#include <math.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, char> pic;
#define bit(x,v) ((ll)x << v)

const ll INF = 1000000007;
const int MAX = 210000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
int pr[100010];
void uini(int n) {
	for (size_t i = 0; i <= n; i++)
	{
		pr[i] = i;
	}
}

int parent(int x) {
	if (x == pr[x]) return x;
	return pr[x] = parent(pr[x]);
}

bool unit(int x, int y) {
	int px = parent(x);
	int py = parent(y);

	if (px == py) return false;
	if (px < py) {
		pr[py] = px;
	}
	else {
		pr[px] = py;
	}
	return true;
}

ll bit[200010];
int max_n = 200000;
int pm = 0;
void add(int x) {

	while (max_n >= x)
	{
		bit[x]++;
		x += x & -x;
	}
}
void sub(int x) {
	while (max_n >= x)
	{
		bit[x]--;
		x += x & -x;
	}
}
ll merge(ll *a, int left, int mid, int right) {
	ll n1 = mid - left;
	ll n2 = right - mid;
	vector<int> L(n1 + 1);
	vector<int> R(n2 + 1);
	for (size_t i = 0; i < n1; i++)
	{
		L[i] = a[left + i];
	}
	for (size_t i = 0; i < n2; i++)
	{
		R[i] = a[mid + i];
	}

	L[n1] = INF;
	R[n2] = INF;
	ll i = 0;
	ll j = 0;
	ll r = 0;
	for (size_t k = left; k < right; k++)
	{
		if (L[i] <= R[j]) {
			a[k] = L[i];
			i++;
		}
		else {
			a[k] = R[j];
			r += n1 - i;
			j++;
		}
	}
	return r;
}
ll mergeSort(ll *a, int left, int right) {
	ll res = 0;
	if (left + 1 < right) {
		int mid = (left + right) / 2;
		res = mergeSort(a, left, mid);
		res += mergeSort(a, mid, right);
		res += merge(a, left, mid, right);
	}
	return res;
}

void solv() {
	int n;
	cin >> n;
	int a[2000100];
	for (size_t i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	int c[10010];
	memset(c, 0, sizeof(c));
	for (size_t i = 0; i < n; i++)
	{
		c[a[i]]++;
	}
	for (size_t i = 1; i < 10010; i++)
	{
		c[i] += c[i - 1];
	}

	int b[2000010];
	memset(b, 0, sizeof(b));
	for (int i = n - 1; i >= 0; i--)
	{
		b[c[a[i]]] = a[i];
		c[a[i]]--;
	}

	for (size_t i = 1; i <= n; i++)
	{
		if (i > 1) cout << " ";
		cout << b[i];
	}
	cout << endl;
}
int main() {
	//	COMinit();
	solv();
	return 0;
}