二分法找中位数

/*Name: 中位数median Author: 巧若拙 Description: 描述给定n（n为奇数且小于1000）个整数，整数的范围在0-m（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。输入第一行是两个整数n和m。第二行是用空格隔开的n个整数 输出一个中位数 样例输入5 103 7 2 5 4 样例输出4*/#include<iostream>  #include<cstdio>  #include<cstdlib>    using namespace std;    const int MAXN = 1000;  int A[MAXN];    int main()   {  int n, m;    cin >> n >> m;      for (int i=0; i<n; i++)          cin >> A[i];          int mid, count, left = 0, right = m;     while (left <= right)      {          mid = (left + right) / 2;          count = 0;        for (int i=0; i<n; i++)        {if (A[i] >= mid)//记录不小于mid的整数数量count++; }        if (count > n/2)                left = mid + 1;          else              right = mid - 1;               //   cout << mid << " " << left << " " << right << " " << count << " " << n/2 << endl;    }            cout << right << endl;             return 0;  }