1324D - Pair of Topics[二分]

time limit per test	memory limit per test	input	output
2 seconds	256 megabytes	standard input	standard output

Description:

The next lecture in a high school requires two topics to be discussed. The i i i-th topic is interesting by a i a_i ai units for the teacher and by b i b_i bi units for the students.
The pair of topics i i i and j ( i < j ) j (i<j) j(i<j) is called good if a i + a j > b i + b j a_i+a_j>b_i+b_j 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 n n ( 2 ≤ n ≤ 2 ⋅ 1 0 5 ) (2≤n≤2⋅10^5) (2≤n≤2⋅105) — the number of topics.
The second line of the input contains n n n integers a 1 , a 2 , … , a n ( 1 ≤ a i ≤ 1 0 9 ) a_1,a_2,…,a_n (1≤a_i≤10^9) a1,a2,…,an(1≤ai≤109), where a i a_i ai is the interestingness of the i i i-th topic for the teacher.
The third line of the input contains n n n integers b 1 , b 2 , … , b n ( 1 ≤ b i ≤ 1 0 9 ) b_1,b_2,…,b_n (1≤b_i≤10^9) b1,b2,…,bn(1≤bi≤109), where b i b_i bi is the interestingness of the i i i-th topic for the students.

Output

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

One Example input

5
4 8 2 6 2
4 5 4 1 3

One Example output

7

Two Example input

4
1 3 2 4
1 3 2 4

Two Example output

0

分析：
题意：
求 a i + a j > b i + b j ( i < j ) a_i + a_j > b_i + b_j(i < j) ai+aj>bi+bj(i<j)的对数
做法：
其实就是求 a i − b i > − ( a j − b j ) ( i < j ) a_i - b_i > -(a_j - b_j)(i < j) ai−bi>−(aj−bj)(i<j)的对数
即 − ( a i + b i ) < a j + b j ( i < j ) -(a_i + b_i) < a_j + b_j (i < j) −(ai+bi)<aj+bj(i<j)的对数
现在令 c i = a i + b i c_i = a_i + b_i ci=ai+bi（方便写）
仔细想一下，如果现在有一个 − c i > c j ( i > j ) -c_i > c_j(i > j) −ci>cj(i>j)其实这个是不符合的
但是将 i , j i, j i,j换一下，其实也就是我们要求的 − c j > c i ( j < i ) -c_j > c_i(j < i) −cj>ci(j<i)
所以其实跟位置并没有什么关系
只要排序一下每次找到大于 − c i -c_i −ci的第一个数的位置就可以了

Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5;int a[maxn];
int b[maxn];int main() {int n;scanf("%d", &n);for(int i = 1; i <= n; ++i)scanf("%d", &a[i]);for(int i = 1; i <= n; ++i)scanf("%d", &b[i]);for(int i = 1; i <= n; ++i)a[i] -= b[i];ll ans = 0;sort(a + 1, a + 1 + n);for(int i = 1; i <= n; ++i) {int x = upper_bound(a + i + 1, a + 1 + n, - a[i]) - a;ans += (n - x + 1) * 1ll;}printf("%lld\n", ans);return 0;
}

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