sicily 1154. Easy sort

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.

 

弱智版的排序题，连输出格式都不用特别注意，直接一堆换行……

如果不自己写排序，用库里自带的qsort函数，那就是神之水题了= =|||||||

一开始还以为要去重，就像弱智版的明明随机数，结果WA，才发现连去重都不要，只要排序再输出就行，果然水……

直接用了以前写的选择排序函数……简单问题复杂化，代码好长（￣へ￣）

 

#include<stdio.h>
#define MAX 1001

void selectionSort( int array[], int size );
void swapInArray( int array[], int first, int second );

int main()
{
    int array[MAX] = {0};
    int t, n;
    int i, j;
    
    scanf( "%d", &t );
    
    for ( i = 0; i < t; i++ )
    {
        scanf( "%d", &n );
        
        for ( j = 0; j < n; j++ )
        {
            scanf( "%d", &array[j] );
        }
        
        selectionSort( array, n );
        
        for ( j = 0; j < n; j++ )
        {
            printf( "%d\n", array[j] );
        }
        
    }
    
    return 0;
}

void selectionSort( int array[], int size )
{
    int i, j;
    int min;
    
    for ( i = 0; i < size - 1; i++ )
    {
        min = i;
        for ( j = i + 1 ; j < size; j++ )
        {
            if ( array[j] < array[min] )
            {
                min = j;
            }
        }
        
        if ( i != min )
        {
            swapInArray( array, i, min );
        }
    }
    
    return;
}


void swapInArray( int array[], int first, int second )
{
    int temp;
    
    temp = array[first];
    array[first] = array[second];
    array[second] = temp;
    
    return;
}