题目描述

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样例

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算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

//对纵坐标y用线段树

#include<iostream>
#include<cstring>
using namespace std;

typedef long long LL;
const int N = 200010;
int a[N], tr[N], lower[N], bigger[N];
int n;

int lowbit(int x)
{
    return x & -x;
}

void add(int x, int c)//修改某个值
{
    for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

int sum(int x)//查询某一段的和
{
    int res = 0;
    for (int i = x; i; i -= lowbit(i)) res += tr[i];
    return res;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> a[i];

    for (int i = 1; i <= n; i ++)
    {
        int y = a[i];
        bigger[i] = sum(n) - sum(y);
        lower[i] = sum(y - 1);
        add(y, 1);
    }

    memset(tr, 0, sizeof tr);
    LL res1 = 0, res2 = 0;//赋初值；

    for (int i = n; i; i --)
    {
        int y = a[i];
        res1 += bigger[i] * (LL)(sum(n) - sum(y));
        res2 += lower[i] * (LL)sum(y - 1);
        add(y, 1);
    }
    cout << res1 << " " << res2 << endl;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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