UVA 11462 Age Sort（計數排序法 優化輸入輸出）

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

 

題目大意：給定若干居民的年齡（都是1-100之間的整數），把他們按照從小到大的順序輸出。

分析：年齡排序。由於資料太大，記憶體限制太緊（甚至都不能把他們全部讀進記憶體），因此無法使用快速排序方法。但整數範圍很小，可以用計數排序法。

程式碼如下：

 

#include<cstdio>
#include<cstring>
#include<cctype> // 為了使用isdigit巨集

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]; // 宣告成全域性變數可以減小開銷
inline void writeint(int i) {
  int p = 0;
  if(i == 0) p++; // 特殊情況：i等於0的時候需要輸出0，而不是什麼也不輸出
  else while(i) {
    buf[p++] = i % 10;
    i /= 10;
  }
  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]); // 逆序輸出
}

int main() {
  int n, x, c[101];
  while(n = readint()) {
    memset(c, 0, sizeof(c));
    for(int i = 0; i < n; i++) c[readint()]++;
    int first = 1;
    for(int i = 1; i <= 100; i++)
      for(int j = 0; j < c[i]; j++) {
        if(!first) putchar(' ');
        first = 0;
        writeint(i);
      }
    putchar('\n');
  }
  return 0;
}