AC码
```
#include<bits/stdc++.h>
using namespace std;
long a[200001];
long n,c,ans;
int main()
{
cin>>n>>c;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
sort(a+1,a+n+1);
for(int i=1;i<=n;i++)
{
ans+=((upper_bound(a+1,a+n+1,a[i]+c)-a)-(lower_bound(a+1,a+n+1,a[i]+c)-a));
}
cout<<ans;
return 0;
}
```
by lorry26 @ 2023-07-04 11:21:22