題目連結：https://www.luogu.com.cn/problem/P3608

方法一

用樹狀陣列求逆序對先後掃兩遍，一次從前往後，一次從後往前，算出每頭奶牛左右兩邊比她高的數量。

最後統計一下。

#include <bits/stdc++.h>

using namespace std;

int sum[500010], l[100010], r[100010];

int n, m, u, v, a[500010], t[500010];

int ans;

inline int read()

{

    int x = 0;

	int f = 1; char ch = getchar();

    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
 

    while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}

    return x * f;

}

int lowbit(int x)

{

	return x & (-x);

}

void add(int i, int v)

{

ㅤㅤㅤㅤ while (i <= n)

	{

		sum[i] += v;

		i += lowbit(i);

	}

}

int query(int i)

{

	int ans = 0;

ㅤㅤㅤㅤ while (i)

	{

		ans += sum[i];
 

		i -= lowbit(i);

	}

	return ans;

}

int main()

{

	n = read();

	for (int i = 1; i <= n; i++)

		a[i] = read(), t[i] = a[i];

ㅤㅤㅤㅤ sort(t + 1, t + 1 + n);

	m = unique(t + 1, t + 1 + n) - t - 1;

	for (int i = 1; i <= n; i++)

		a[i] = lower_bound(t + 1, t + 1 + m, a[i]) - t;

ㅤㅤㅤㅤ for (int i = 1; i <= n; i++)
 

	{

		add(a[i], 1);

		l[i] = i - query(a[i]);

	}

	memset(sum, 0, sizeof sum);

ㅤㅤㅤㅤ for (int i = n; i >= 1; i--)

	{

		add(a[i], 1);

		r[i] = n - i + 1 - query(a[i]);

	}

ㅤㅤㅤㅤ for (int i = 1; i <= n; i++)

		if (max(l[i], r[i]) > 2 * min(l[i], r[i]))

			ans++;

	printf("%d\n", ans);

	return 0;

}