EOJ When I Sort...

EOJ When I Sort…
In datastruct class, the teacher gives partychen a simple sort question:A permutation $p[0]$, $p[1]$, …, $p[n-1]$ is a sequence containing each number from $0$ to $n-1$ exactly once. The result of applying permutation $p$ to an array $A$ of length $n$ is an array $B$ of length $n$, where $B[p[i]]=A[i]$ (0-based indices).
Given an array $A$, find a permutation which has the effect of sorting the elements of $A$ in non-descending order, i.e., an order where each element is greater than or equal to the previous one. If there are several suitable permutations output the lexicographically smallest one.
The permutation $p[0]$, $p[1]$, …, $p[n-1]$ is considered lexicographically smaller than the permutation $q[0]$, $q[1]$, …, $q[n-1]$ if there is an index $i$ such that $p[i]<q[i]$ and the equation $p[j]=q[j]$ holds for all $j<i$.

输入格式
The first line of input gives the number of cases, $N$($1\le N\le 100$). $N$ test cases follow.
In each test case, there are two lines. The first line is a interger $M$($1\le M\le 10000$), representing the numbers of array $A$, and in the second line there are $M$ intergers.

输出格式
Output the permutation $p$ as discrible above.

样例输入
2
3
2 3 1
4
2 1 3 1

输出：
1 2 0
2 0 3 1

#include <iostream>
#include <string.h>
#include <math.h>
#include <map>
#include <vector>
#include <sstream>
#include <algorithm>
#include <iomanip>
using namespace std;
struct t {
	int index;
	int num;
}a[10005];
int cmp(t aa, t bb)
{
	if (aa.num == bb.num)
		return aa.index < bb.index;
	return aa.num < bb.num;

}
int main()
{
	int n;
	cin >> n;
	int arr[10005] = { 0 };
	while (n--)
	{
		
		int m; cin >> m;
		for (int i = 0; i < m; i++)
		{
			cin >> a[i].num;
			arr[i] = a[i].num;
			a[i].index = i;
		}
		sort(a, a + m, cmp);
		for (int i = 0; i < m; i++)
		{
			for (int j = 0; j < m; j++)
			{
				if (arr[i] == a[j].num && i == a[j].index)
					cout << j << ' ';
			}
		}
		cout << endl;
	}
}