[SOJ]Easy sort （归并排序）

Description

You know sorting is very important. And this easy problem is:

Given you an array with N non-negative integers which are smaller than 10,000,000, you have to sort this array. Sorting means that integer with smaller value presents first.

Input

The first line of the input is a positive integer T. T is the number of the test cases followed.

The first line of each test case is a positive integer N (1<= N<= 1000) which represents the number of integers in the array. After that, N lines followed. The i-th line is the i-th integer in the array.

Output

The output of each test case should consist of N lines. The i-th line is the i-th integer of the array after sorting. No redundant spaces are needed.

Sample Input

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2
3
1
2
3
1
1

Sample Output

1
2
3
1

#include<cstdio>
#define MAX 1001
int a[MAX], aux[MAX];

//2路归并排序
void merge_sort(int lo, int hi)
{
    if (lo < hi)
    {
        int mid = lo + (hi - lo)/2;
        merge_sort(lo, mid);  //递归
        merge_sort(mid+1, hi);

        for (int i = lo; i <= hi; ++i)   //赋初值
            aux[i] = a[i];

        int l = lo, r = mid+1;
        for (int i = lo; i <= hi; i++)
        {
            if (l > mid) a[i] = aux[r++];    //第一路已经处理完
            else if (r > hi) a[i] = aux[l++];   //第二路已经处理完
            else if (aux[r] < aux[l]) a[i] = aux[r++];   //二路比较
            else a[i] = aux[l++];
        }
    }
}

int main()
{
    int t, n;

    scanf("%d", &t);

    while(t--)
    {
        scanf("%d", &n);

        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]);

        merge_sort(0, n-1);

        for (int i = 0; i < n; ++i)
            printf("%d\n", a[i]);
    }

    return 0;
}