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
思路:计数排序
AC代码1(519ms):
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { int x, n, c[101]; while(scanf("%d", &n) == 1 && n) { memset(c, 0, sizeof(c)); for(int i=0; i<n; i++) { scanf("%d", &x); c[x]++; } int flag = 1; for(int i=1; i<=100; i++) for(int j=0; j<c[i]; j++) { if(!flag) printf(" "); flag = 0; printf("%d", i); } printf("\n"); } return 0; }
AC代码2(172ms):
#include <cstdio> #include <cstring> #include <algorithm> #include <cctype> //为了使用isdigit宏 using namespace std; 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]; //声明成全局变量可以减小开销 inline void writeint(int i) { int p = 0; if(i == 0) 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 % 10; i /= 10; } for(int j = p-1; j >= 0; j--) putchar('0' + buf[j]); //逆序输出 } 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 flag = 1; for(int i = 1; i <= 100; i++) for(int j=0; j < c[i]; j++) { if(!flag) putchar(' '); flag = 0; writeint(i); } putchar('\n'); } return 0; }