Age Sort UVA - 11462(计数排序）

题目链接

https://vjudge.net/problem/UVA-11462

题目

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

题意

给定n个居民的年龄（1~99），把它们按照由小到大的顺序输出。

分析

如果直接用sort排序，需要用一个数组来存放n个居民的年龄，而题目中说明输入有25M，内存限制为2M。

所以理论上不能直接调用sort。（但我直接用sort也AC了。。。）

虽然n很大，但是每个居民的年龄范围很小。所以，可以考虑用计数排序。把每个年龄的人数纪录下来，然后按照年龄和该年龄的人数输出即可。这样只需开一个大小为101的数组。

代码

#include <cstdio>
#include <cstring>

using namespace std;

int n,buf[101];

int main()
{
while(~scanf("%d",&n)  && n)
{
int x;
memset(buf,0,sizeof(buf));
for(int i=0;i<n;i++)
{
scanf("%d",&x);
buf[x]++;
}
int cnt=0;
for(int i=1;i<=99;i++)
for(int j=1;j<=buf[i];j++)
{
cnt++;
if(cnt==n)
printf("%d\n",i);
else
printf("%d ",i);
}
}
return 0;
}