Input

The first line is a positive integer t for the number of test cases.
Each test case contains two lines. The first line is an integer n (0<n<=100). The second line contains n integers.

Output

For each test case, outputs the distinct numbers in order in which they were read, seperated by one blank.

Sample Input

2
3
1 2 1
4
2 1 2 1
Sample Output

1 2
2 1

循环判断，应该有分支结构

还是不明白，是要用嵌套数组吗？

你得找到这样一个数据结构：它能快速的新增一个元素，它能快速的判断是否存在某个元素。

#include <stdio.h>
#include <set> // 只是示例

int main( void )
{
    unsigned t;
    scanf( "%u", &t );
    while( t-- )
    {
        std::set<int> container; // 只是示例，假设你有这样的一个数据结构

        unsigned n;
        scanf( "%u", &n );
        while( n-- )
        {
            int v;
            scanf( "%d", &v );

            if( container.find(v) == container.end() ) // 只是示例，判断v是否已经存在于container中
            {
                printf( "%s%d", (" "+container.empty()), v ); // 只是示例，判断container容器是否为空
                container.insert( v ); // 只是示例，将v添加到container容器中
            }
        }
        putchar( '\n' );
    }

    return 0;
}

从题目来看，这是一道比较弱的题，自然不需要你写一个高效的算法，因此循环查找即可。

#include <stdio.h>

int main( void )
{
    unsigned t;
    scanf( "%u", &t );
    while( t-- )
    {
        int container[100];
        size_t container_size = 0;

        unsigned n;
        scanf( "%u", &n );
        while( n-- )
        {
            int v;
            scanf( "%d", &v );

            size_t i;
            for( i=0; i!=container_size && v!=container[i]; ++i );

            if( i == container_size )
            {
                printf( "%s%d", (" "+(container_size==0)), v );
                container[container_size++] = v;
            }
        }
        putchar( '\n' );
    }

    return 0;
}