224. NOIP2015PJ-28（题解）

题面

28．（完善程序）

（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 至 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。（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(_________________________)
    {
        mid = (lbound + rbound) / 2;
        _________________________;
        for(i = 0; i < n; i++)
            if(_________________________)
            _________________________;
        if(count > n / 2)
            lbound = mid + 1;
        else
            _________________________;
        cout << mid << " " << lbound << " " << rbound << " " << count << endl;
    }
    cout << rbound << endl;
    return 0;
}

分析

1.二分法模板（直接套）

秒了1，2，4，5题好吧~

2.思维

注意！mid是一个数，要用mid与x数组的数来比较 而不是x数组与x数组的数进行比较

3.解析

①：[lbound<=rbound]

②：[count=0]

③：[mid<=x[i]]

④：[count++]

⑤：[rbound=mid-1]

最后！！！

不要输入多余的空格！（只写代码）

吃过亏的都emo......了

①： lbound<=rbound

②： count=0

③： mid<=x[i]

④： count++

⑤： rbound=mid-1

#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(_________________________)
    {
        mid = (lbound + rbound) / 2;
        _________________________;
        for(i = 0; i < n; i++)
            if(_________________________)
            _________________________;
        if(count > n / 2)
            lbound = mid + 1;
        else
            _________________________;
        cout << mid << " " << lbound << " " << rbound << " " << count << endl;
    }
    cout << rbound << endl;
    return 0;
}