Divide and conquer:Median(POJ 3579)

　　　　　　　　　　　     快速求两数距离的中值

　　题目大意：给你一个很大的数组，要你求两个数之间的距离的中值

　　二分法常规题，一个pos位就搞定的事情

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <functional>
 4
 5 using namespace std;
 6 typedef long long LL_INT;
 7
 8 static int nums[100002];
 9 bool judge(LL_INT, LL_INT,const int);
10
11 int main(void)
12 {
13     int lb, rb, mid;
14     LL_INT k, n;
15
16     while (~scanf("%lld", &n))
17     {
18         k = n*(n - 1) / 2;
19         k = k % 2 == 0 ? k / 2 : k / 2 + 1;
20
21         for (int i = 0; i < n; i++)
22             scanf("%d", &nums[i]);
23         sort(nums, nums + n);
24         lb = 0; rb = 1000000001;
25         while (rb - lb > 1)
26         {
27             mid = (lb + rb) / 2;
28             if (judge(k, n, mid)) lb = mid;
29             else rb = mid;
30         }
31         printf("%d\n", rb);
32     }
33     return 0;
34 }
35
36 bool judge(LL_INT k, LL_INT n,const int x)
37 {
38     int i = 0, sum = 0, pos = 1;
39     for (; i < n; i++)
40     {
41         for (; pos < n && nums[pos] - nums[i] <= x; pos++);
42         sum += pos - i - 1;
43         if (sum >= k)
44             return false;
45     }
46     return true;
47 }