Age Sort

Input: Standard Input

Output: Standard Output

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 Output for Sample Input

3 4 2 1 5

2 3 2 3 1

1 2 3 4 5

1 2 2 3 3

Note: The memory limit of this problem is 2 Megabyte Only.

本题题意：

给定若干居民的年龄（1-100之间的整数），把他们按照从小到大的顺序输出。

输入包含多组测试数据，第一行一个整数n（0<n<=2000 000），即居民总数，下一行包含n个表示年龄的整数，输入结束标志：n = 0.

输入文件约有25MB，而内存限制只有2MB。

分析：

由于数据太大，内存限制太紧，根本无法使用快速排序，因此可以使用计数排序（桶排序）的方法。

#include<cstdio>
#include<cstring>   //使用memset
using namespace std;

int main()
{
    int n , num[101];
    while(scanf("%d" , &n) == 1 && n)
    {
        int i , j , temp;
        memset(num , 0 , sizeof(num));
        for(i = 0; i < n; i++)
        {
            scanf("%d" , &temp);
            num[temp]++;
        }
        int first = 1;
        for(i = 1; i <= 100; i++)
        {
            for(j = 0; j < num[i]; j++)
            {
                if(!first) printf(" ");
                first = 0;
                printf("%d" , i);
            }
        }
        printf("\n");
    }
    return 0;
}

View Code

还有一种更加节约时间的代码，当输入的数据量很大时，使用printf和scanf函数，如果还不行，就要使用字符一个一个的输入。

由于使用字符的形式一个一个的输入，所以将函数定义为内联函数。
解题代码如下：

#include<cstdio>
#include<cctype>  //使用isdigit宏的头文件
#include<cstring>
using namespace std;

inline int readin()   //使用内联函数
{
    int c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c))
    {
        x = x * 10 + (c - '0');
        c = getchar();
    }
    return x;
}
inline void writeout(int n)
{
    int i = 0 , j;
    int temp[3];
    if(n == 0)
    {
        temp[0] = 0;
        i++;
    }
    else
    {
        while(n)
        {
            temp[i++] = n % 10;
            n /= 10;
        }
    }
    for(j = i-1; j >=0; j--) putchar('0' + temp[j]);
}
int main()
{
    int n , num[101];
    //freopen("data.txt","r",stdin);//读
    while(n = readin())
    {
        int i , j;
        memset(num , 0 , sizeof(num));
        for(i = 0; i < n; i++)
        {
            num[readin()]++;
        }
        int first = 1;
        for(i = 1; i <= 100; i++)
        {
            for(j = 0; j < num[i]; j++)
            {
                if(!first) putchar(' ');
                first = 0;
                writeout(i);
            }
        }
        putchar('\n');
    }
    return 0;
}

View Code