二分查找 原题在洛谷p1102 A-B数对

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll arr[200000];
ll n,c;
ll my_lower_bound(ll x){//该函数的目的是找到数组中第一个不小于x的下标值 
	ll left=0,right=n-1;
	//利用二分搜索，复杂度log(n),非常高效  arr数组已经升序 
	while(left<=right){
		ll mid=left+(right-left)/2;
		if(arr[mid]<x){
			left=mid+1;
		}
		else{
			right=mid-1;
		}
	}
	return left;
}
ll my_upper_bound(ll x){//该函数的目的是找到数组中第一个大于x的下标值 
	ll left=0,right=n-1;
	//利用二分搜索，复杂度log(n),非常高效  arr数组已经升序 
	while(left<=right){
		ll mid=left+(right-left)/2;
		if(arr[mid]>x){
			right=mid-1;
		}
		else{
			left=mid+1;
		}
	}
	return left;
}
ll sum(ll x){
	ll n1=my_lower_bound(x);
	if(arr[n1]!=x){
		return 0;
	}
	ll n2=my_upper_bound(x);
	return n2-n1;
}
void solve(){
	ll ret=0;
	for(ll i=0;i<n;i++){
		ret+=sum(arr[i]+c);
	}
	cout<<ret;
}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n>>c;
	for(ll i=0;i<n;i++){
		cin>>arr[i];
	}
	sort(arr,arr+n);
	solve();
	return 0;
}