POJ-3579 二分答案

题目链接: http://poj.org/problem?id=3579

题意: 通俗易懂

思路: 二分答案即可

总结: 基本的二分操作

代码如下:

 1 import java.util.*;
 2 
 3 public class Main {
 4     public static int[] a;
 5 
 6     public static int n;
 7     public static int k;
 8     public static long an;
 9 
10     public static void main(String[] args) {
11         Scanner sc = new Scanner(System.in);
12         while (sc.hasNext()) {
13             n = sc.nextInt();
14             an = ((long) n * (n - 1)) >> 1;
15             an = (an + 1) >> 1;
16             a = new int[n];
17             for (int i = 0; i < n; ++i) {
18                 a[i] = sc.nextInt();
19             }
20             Arrays.sort(a);
21             int left = -1;
22             int right = a[n - 1] - a[0] + 1;
23             while (right - left > 1) {
24                 int mid = (left + right) >> 1;
25                 if (C(mid)) {
26                     right = mid;
27                 } else
28                     left = mid;
29             }
30             System.out.println(right);
31         }
32     }
33 
34     public static boolean C(int mid) {
35         int j = 0;
36         int count = 0;
37         for (int i = 1; i < n; i++) {
38             while (a[i] - a[j] > mid)
39                 j++;
40             count += (i - j);
41         }
42         return count >= an;
43     }
44 }