中位数

median) 给定 n(n 为奇数且小于 1000)个整数，整数的范围在 $(0<m<2^{31})$ 之间，请使用二分法求这 n 个整数的中位数。所谓中位数，是指将这 n 个数排序之后，排在正中间的数。(第五空 2 分，其余 3 分)

#include <iostream>
using namespace std;
const int MAXN = 1000;
int n, i, lbound, rbound, mid, m, count;
int x[MAXN];
int main()
{
    cin >> n >> m;
    for (i = 0; i < n; i++)
        cin >> x[i];
    lbound = 0;
    rbound = m;
    while (①)
    {
        mid = (lbound + rbound) / 2;
        ②;
        for (i = 0; i < n; i++)
            if (③)
                ④;
        if (count > n / 2)
            lbound = mid + 1;
        else
            ⑤;
        cout << mid << " " << lbound << " " << rbound << " " << count << endl;
    }
    cout << rbound << endl;
    return (0);
}

1. ①处应填（ ）{{ select(1) }}

• lbound<rbound
• lbound+1<rbound
• lbound<=rbound
• rbound - lbound > 1

2. ②处应填（ ）{{ select(2) }}

• int count=0
• int p=0
• int count=1
• int p=1

3. ③处应填（ ）{{ select(3) }}

• x[i]>x[p]
• x[i]>=x[p]
• mid<x[i]
• x[mid]>i

4. ④处应填（ ）{{ select(4) }}

• p=i
• p=max(p,i)
• count*=1
• count++

5. ⑤处应填（ ）{{ select(5) }}

• mid=rbound
• rbound=mid+1
• rbound=mid
• rbound=mid-1