UVA 11462 Age Sort（计数排序法 优化输入输出）

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

Output for Sample Input

1 2 3 4 5

1 2 2 3 3

题目大意：给定若干居民的年龄（都是1-100之间的整数），把他们按照从小到大的顺序输出。

分析：年龄排序。由于数据太大，内存限制太紧（甚至都不能把他们全部读进内存），因此无法使用快速排序方法。但整数范围很小，可以用计数排序法。

代码如下：

 #include<cstdio>
 #include<cstring>
 #include<cctype> // 为了使用isdigit宏

 inline int readint() {
   char c = getchar();
   while(!isdigit(c)) c = getchar();

   int x = ;
   while(isdigit(c)) {
     x = x *  + c - '';
     c = getchar();
   }
   return x;
 }

 int buf[]; // 声明成全局变量可以减小开销
 inline void writeint(int i) {
   int p = ;
   if(i == ) p++; // 特殊情况：i等于0的时候需要输出0，而不是什么也不输出
   else while(i) {
     buf[p++] = i % ;
     i /= ;
   }
   for(int j = p-; j >=; j--) putchar('' + buf[j]); // 逆序输出
 }

 int main() {
   int n, x, c[];
   while(n = readint()) {
     memset(c, , sizeof(c));
     for(int i = ; i < n; i++) c[readint()]++;
     int first = ;
     for(int i = ; i <= ; i++)
       for(int j = ; j < c[i]; j++) {
         if(!first) putchar(' ');
         first = ;
         writeint(i);
       }
     putchar('\n');
   }
   return ;
 }