shǎ崽 OrOrOrOrz
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7821 Accepted Submission(s): 3735
Problem Description
Acmer in HDU-ACM team are ambitious, especially shǎ崽, he can spend time in Internet bar doing problems overnight. So many girls want to meet and Orz him. But Orz him is not that easy.You must solve this problem first.
The problem is :
Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
For example, give you 1 2 3 4 5, you should output 5 1 4 2 3
The problem is :
Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
For example, give you 1 2 3 4 5, you should output 5 1 4 2 3
Input
There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.
Output
Output a sequence of distinct integers described above.
Sample Input
5 1 2 3 4 5
Sample Output
5 1 4 2 3
Author
WhereIsHeroFrom
Source
Recommend
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 using namespace std; 6 int num[10010]; 7 int main() 8 { 9 int n; 10 while(~scanf("%d", &n)) 11 { 12 for(int i = 0; i < n; i++) 13 scanf("%d", &num[i]); 14 sort(num, num + n); 15 if(n & 1) 16 { 17 for(int j = 0, i = n-1; i > n/2; j++, i--) 18 printf("%d %d ", num[i], num[j]); 19 printf("%d ", num[n/2]); 20 } 21 else 22 { 23 for(int j = 0, i = n-1; i >= n/2; j++, i--) 24 { 25 if(j == 0) 26 printf("%d %d", num[i], num[j]); 27 else 28 printf(" %d %d", num[i], num[j]); 29 } 30 printf(" "); 31 } 32 } 33 34 return 0; 35 }