28| A-B数对

代码实现

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 2e5 + 10; 
unordered_map <int, int> a;
int cnt, n, c, b[N];

signed main()
{
	cin >>  n >> c;
	for (int i = 1; i <= n; i++) 
	{
		cin >> b[i];
		a[b[i]]++;
	}
	for (int i = 1; i <= n; i++)
	{
		cnt += a[b[i] + c];
	}
	cout << cnt << endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 2e5 + 10; 
LL cnt, n, c, b[N];

int main()
{
	cin >>  n >> c;
	for (int i = 1; i <= n; i++) cin >> b[i];
	// 预处理 有序 
	sort(b+1, b+n+1);
	for (int i = 1; i <= n; i++)
	{
		LL x = b[i] + c;
		auto l = lower_bound(b + 1, b + n + 1, x);
		auto r = upper_bound(b + 1, b + n + 1, x); 
		cnt += (r - l);
	}
	cout << cnt << endl;
	return 0;
}