UVA - 11462 Age Sort

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 le. 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

分析：

刘汝佳大白书里的题，因为数据太大，内存限制太紧，但整数范围很小，可以用计数排序法。

解法一：

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100+5;
int n,a[maxn];
int main()
{
    while(scanf("%d",&n)==1&&n)
    {
        int x;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            a[x]++;
        }
        bool flag=true;
        for(int i=1;i<=100;i++)
        for(int j=1;j<=a[i];j++)
        {
            if(!flag)
            printf(" ");
            flag=false;
            printf("%d",i);
        }
        printf("\n");
    }
    return 0;
}

解法二：（如果时间卡的更紧的话，此方法使时间缩短约2/3）

优化输入输出，还是挺有趣的，以后如果遇到卡IO的题可以用这种方法。

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cctype>
using namespace std;
const int maxn=100+5;
int n,a[maxn];


inline int readint()
{
    char c=getchar();
    while(!isdigit(c)) c=getchar();
    int x=0;
    while(isdigit(c)) 
    {
        x=x*10+c-'0';
        c=getchar();
    }
    return x;
}


int buf[10];
inline void writeint(int i)
{
    int p=0;
   // if(i==0) p++;//对于这道题，这两行可以不要。
    //else
    while(i)
    {
        buf[p++]=i%10;
        i/=10;
    }
    for(int j=p-1;j>=0;j--)
    putchar('0'+buf[j]);
}


int main()
{
    while(n=readint())
    {
        int x;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
           a[readint()]++;
        }
        bool flag=true;
        for(int i=1;i<=100;i++)
        for(int j=1;j<=a[i];j++)
        {
            if(!flag)
            putchar(' ');
            flag=false;
            writeint(i);
        }
        putchar('\n');
    }
    return 0;
}