标签：style blog http io color ar os java for
题意：一个1~n的无重复整数序列，要求打印其中一种排列，使新序列相邻两项之差的绝对值去重后的个数刚好为k个
(1 <= k < n <= 10^5)
输入：n,k
输出：新序列
思路：想了一下，先考虑k最大和k最小取值时的序列排列情况
1) k最大时：k = n-1 序列：1,n,2,n-1,3,n-2.... 相邻两项绝对值差：n-1,n-2,n-3...1
2) k最小时：k = 1 序列：1,2,3...n | n,n-1,n-2...1
k取值为 [1,n-1] 那么中间的取值就这两个情况进行组合，前段用1)，剩下数用2)
当k为奇数时，2)用升序；当k为偶数时，2)用降序
好，撸代码
1 package golao.org.cf275; 2 3 import java.io.BufferedReader; 4 import java.io.IOException; 5 import java.io.InputStreamReader; 6 import java.util.Arrays; 7 import java.util.StringTokenizer; 8 9 public class DiversePermutation { 10 11 public static void main(String[] args) throws IOException { 12 BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 13 StringTokenizer stk = new StringTokenizer(br.readLine()); 14 int n,k; 15 n = Integer.parseInt(stk.nextToken()); 16 k = Integer.parseInt(stk.nextToken()); 17 System.out.println(print(findPermutation(n,k))); 18 19 } 20 public static int[] findPermutation(int n,int k) 21 { 22 int[] ary = new int[n]; 23 for(int i=1;i<=n;i++){ 24 ary[i-1] = i; 25 } 26 int[] aryResult = new int[n]; 27 int count = n-1; 28 //前段 29 for(int j=1,i=0;j<=k;j+=2,i++){ 30 aryResult[j-1] = ary[i]; 31 aryResult[j] = ary[count--]; 32 } 33 //后段 34 if(k%2==0){ 35 for(int index=k;index<n;index++){ 36 aryResult[index] = ary[count--]; 37 } 38 }else{ 39 for(int index=k;index<n;index++){ 40 aryResult[index] = aryResult[index-1]+1; 41 } 42 } 43 return aryResult; 44 } 45 public static String print(int[] ary){ 46 StringBuffer sb = new StringBuffer(""); 47 for(int i=0;i<ary.length;i++){ 48 sb.append(ary[i]); 49 sb.append(" "); 50 } 51 return sb.toString(); 52 } 53 54 }
提交给CF，运行通过，77ms
CF 275(div.1) A Diverse Permutation
标签：style blog http io color ar os java for
原文地址：http://www.cnblogs.com/golao/p/4071215.html
