84,二分，TLE

#include<bits/stdc++.h>
using namespace std;
const int N=220000;
int sum=0;
long long a[N];
int read(){
    int x=0,w=1;
    char ch=0;
    while(ch<'0'||ch>'9'){
        if(ch=='-')w=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+(ch-'0');
        ch=getchar();
    }
    return x*w;
}//快读不用管； 
int main() {
	int n;
    long long int c;
	n=read();
	c=read();
	for(int i=1; i<=n; ++i)a[i]=read();
	sort(a+1,a+n+1);
	for(int i=1; i<=n; ++i){
		if(a[i]<c)continue;//a=b+c,c>=1,a[i]必须大于c 
		else {
			int l=1,r=i-1;
			while(l<r) {
				int mid=(l+r)>>1;
				if(a[mid]>=a[i]-c)r=mid;
				else l=mid+1;
			}//左边界 
			if(a[l]!=a[i]-c)continue;//没有这个数就继续下一个； 
			else {
				int l1=1,r1=i-1;
				while(l1<r1) {
					int mid=(l1+r1+1)>>1;
					if(a[mid]<=a[i]-c)l1=mid;
					else r1=mid-1;
				}//右边界 
				for(int j=l;j<=l1;++j){
				        sum+=1;
                    }
                }
			}
		}
		printf("%d",sum);
		return 0;
}//TLE