C++求众数

#include
using namespace std;
const int N = 101;
int a[N];
int sum[N];
int position[N];
void quick_sort(int l, int r, int q[],int p[])
{
	if (l >= r) return;
	int i = l - 1, j = r + 1, x = q[(l + r) >> 1];
	while (i < j)
	{
		do i++; while (q[i] < x);
		do j--; while (q[j] > x);
		if (i < j) {
			swap(q[i], q[j]);
			swap(p[i], p[j]);
		}
	}
	quick_sort(l, j, q,p);
	quick_sort(j + 1, r, q,p);
}
int main()
{
	sum[0] = 1;
	int n,flag=1;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		position[i] = i;
	}
	quick_sort(0, n - 1, a,position);
	for (int i = 0; i < n; i++)
	{
		if (a[i + 1] == a[i]) sum[i + 1] = sum[i]+1;
		else sum[i + 1] = 1;
		position[i] = i;
	}
	quick_sort(0, n - 1, sum, position);
	for (int i = n - 1; i > 0; i--)
	{
		if (sum[i] != sum[i - 1]) break;
		flag++;
	}
	if (flag < n)
		cout << "该数列的众数有" << flag << "个" << endl;
	else {
		cout << "该数列没有众数";
		return 0;
	}
	for (int i =n-1;flag>0; i--,flag--)
	{
		cout << "众数为：" << a[position[i]]<<",";
		cout << "其出现的次数为：" << sum[i]< 

 #时间复杂性：快排采用分治与递归思想，平均时间复杂度为O(nlog2n)，输入比较算法时间复杂度为O(n)，该算法的时间复杂度为O(nlog2n)。