Day3-O-Median POJ3579

Given N numbers, X₁, X₂, ... , X_N, let us calculate the difference of every pair of numbers: ∣X_i - X_j∣ (1 ≤ i ＜ j ≤ N). We can get C(N,2) differences through this work, and now your task is to find the median of the differences as quickly as you can!

Note in this problem, the median is defined as the (m/2)-th  smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

Input

The input consists of several test cases.
In each test case, N will be given in the first line. Then N numbers are given, representing X₁, X₂, ... , X_N, ( X_i ≤ 1,000,000,000  3 ≤ N ≤ 1,00,000 )

Output

For each test case, output the median in a separate line.

Sample Input

4
1 3 2 4
3
1 10 2

Sample Output

1
8

思路：直接N^2肯定行不通，那么我们可以二分差值，满足单调且可验证，最小最大问题，代码如下：

const int maxm = ;

int n, buf[maxm], m;

bool check(int d) {
    int sum = ;
    for (int i = ; i < n; ++i) {
        sum += upper_bound(buf + i, buf + n, buf[i] + d) - buf - i - ;
    }
    return sum >= m;
}

int main() {
    while(scanf("%d",&n) != EOF) {
        for (int i = ; i < n; ++i)
            scanf("%d", &buf[i]);
        sort(buf, buf + n);
        m = n * (n - ) / ;
        if(m %  == )
            m = m / ;
        else
            m = m /  + ;
        int l = , r = buf[n - ] - buf[], mid;
        while(l <= r) {
            mid = (l + r) >> ;
            if(check(mid))
                r = mid - ;
            else
                l = mid + ;
        }
        printf("%d\n", l);
    }
    return ;
}