ACM Age Sort排序

Description

 Age Sort：You are given the ages (in years) of all people of a country with at least 1 year of age. You know that no individual in that country lives for 100 or more years. Now, you are given a very simple task of sorting all the ages in ascending order.

Input

There are multiple test cases in the input file. Each case starts with an integer n (0 < n ≤ 2000000), the total number of people. In the next line, there are n integers indicating the ages. Input is terminated with a case where n = 0. This case should not be processed.

Output

For each case, print a line with n space separated integers. These integers are the ages of that country sorted in ascending order. Warning: Input Data is pretty big (∼ 25 MB) so use faster IO.

Sample Input

5

3 4 2 1 5

5

2 3 2 3 1

0

Sample Output

1 2 3 4 5

1 2 2 3 3

解题思路：首先确定程序循环条件，然后然按照题意先输入n个数字，并把它存在一个数组中，再用sort函数将它排序（sort函数存在于头文件<algorithm>中，它是按从小到大的顺序排序的）排 好序之                  后就要用for循环语句将它输出来，这时要注意空格的输出，可以这样控制空格的输出（首先给定一个first,并将它赋值为1，如果first为1就把它赋值为0；否则就输出空格， 这样就可以保证不                  会输出多余的空格。）

程序代码：

#include <stdio.h>

#include <algorithm>

using namespace std;

const int maxn=2000000;

int a[maxn];

int main ()

{

      int n;

    while(scanf("%d",&n)&&n)

      {

            int first=1;

            for(int i=0;i<n;i++)

                  scanf("%d",&a[i]);

            sort(a,a+n);

            for(int k=0;k<n;k++)

            {

                  if(first)

                        first=0;

                  else

                        printf(" ");

                  printf("%d",a[k]);

            }

            printf("
");

      }

      return 0;

}