Codeforces Round #627 (Div. 3) D - Pair of Topics

**Pair of Topics **

题目描述

The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by ai units for the teacher and by bi units for the students.

The pair of topics i and j (i<j) is called good if ai+aj>bi+bj (i.e. it is more interesting for the teacher).

Your task is to find the number of good pairs of topics.

Input

The first line of the input contains one integer n (2≤n≤2⋅105) — the number of topics.

The second line of the input contains n integers a1,a2,…,an (1≤ai≤10^9), where ai is the interestingness of the i-th topic for the teacher.

The third line of the input contains n integers b1,b2,…,bn (1≤bi≤109), where bi is the interestingness of the i-th topic for the students.

Output

Print one integer — the number of good pairs of topic.

Examples
Input #1
5
4 8 2 6 2
4 5 4 1 3
Output #1
7
Input #2
4
1 3 2 4
1 3 2 4
Output #2
0

Solution

运用了树状数组求逆序对的思想。令ci = ai - bi，原式移项得到
ai - bi > bj - aj (i < j)
即 ci > - cj (i < j)
我们发现与逆序对很相似，只是多了一个负号，为了处理负数和大数，我们需要进行离散化，要比较ci和-cj的大小，所以把ai - bi和bi - ai 一起离散化，然后每次查询已插入的小于等于bi - ai的（a - b）数量，当前是已经插入i - 1 个，故i - 1 - 查询的值即为比bi - ai大的（a - b）数量，计入答案中，然后插入ai - bi。
正向扫一遍最后得到答案。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;

const int SZ = 500000 + 20;
const int INF = 0x3f3f3f3f;

ll ans;
ll tree[SZ],n,a[SZ],b[SZ],all[SZ];

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

inline void update(int x)
{
	while(x <= 2 * n)
	{
		tree[x] += 1;
		x += lowbit(x);
	}
}

inline ll query(int x)
{
	ll sum = 0;
	while(x)
	{
		sum += tree[x];
		x -= lowbit(x);
	}
	return sum;
}

int main()
{
	memset(tree,0,sizeof(tree));
	scanf("%d",&n);
	for(int i = 1;i <= n;i ++ ) scanf("%lld",&a[i]);
	for(int j = 1;j <= n;j ++ ) scanf("%lld",&b[j]); 
	for(int i = 1;i <= n;i ++ )
	{
		ll temp = a[i];
		a[i] = a[i] - b[i];
		b[i] = b[i] - temp;
		all[i] = a[i];
		all[i + n] = b[i];
	}
	sort(all + 1,all + 2 * n + 1);
	int sum = unique(all + 1,all + 2 * n + 1) - all - 1;
	ans = 0;
	for(int i = 1;i <= n;i ++ )
	{
		a[i] = lower_bound(all + 1,all + 1 + sum,a[i]) - all;
		b[i] = lower_bound(all + 1,all + sum + 1,b[i]) - all;
	}
	for(int i = 1;i <= n;i ++ )
	{	
		ans += (i - 1 - query(b[i]));
		update(a[i]);
	}
	printf("%lld",ans);
	return 0;
}

2020.3.17