11462 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：首先是以0为结束标志，所以在输入一个数字之后判断是否为0，再继续进行操作。简单说就是一个排序问题，输入一组数据经过排序后输出，但是这里要注意一个时间问题，如果用cin和cout，那么将很容易时间超出，最好用scanf和printf来进行输入和输出操作，这样才能保证时间在规定的时间内。
程序代码：
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=2000000;
int a[maxn];
int main()
{
int n,p=1;
while(scanf("%d",&n)==1&&n)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int z=0;z<n;z++)
if(a[z]<1||a[z]>=100) { p=0; break;}
else continue;
if(p==1)
{
sort(a,a+n);
for(int x=0;x<n-1;x++)
printf("%d ",a[x]);
printf("%d",a[n-1]);
printf("
");
}
}
return 0;
}