这道题目思路如下。首先把$x$和$y$分开来算。如何算，可以把所有的减去$x$的最小值$xmi$，那么对差值求和就是它在最小值的时候的值。然后考虑移动。向右移动 则左边所有点长度$+1$，右边，所有的点长度$-1$，可以通过循环把从最小值到最大值之间所有的值算出来。由于$D$小于$1e6$ 那么最远是$-2e6$ 和$+2e6$ 因为最小值向左和最大值向右都是增加，所以可以大于$d$的时候就break。对于每个符合的放到桶里面，值++；对桶求前缀和 那么从小到大遍历$x$ 对于$x$的每个值$i$ 结果加上$qzhytong[D-i]$ 这需要注意一点 $D-i$要$>=0$

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n, D;
int x[2000005], y[2000005], xj[4000005], yj[4000005], qzhx[200005], qzhy[200005], xzhi[4000005], yzhi[4000005], xtong[1000005], ytong[1000005], xtongqzh[1000005], ytongqzh[1000005];
signed main() {
//	freopen("02_handmade_01.in","r",stdin);
	cin >> n >> D;
	int xmi = 10000000, ymi = 10000000, xma = -1, yma = -1;


	for (int i = 1; i <= n; i++) {

		cin >> x[i] >> y[i];
		xj[x[i] + 2000000]++;
		yj[y[i] + 2000000]++;
		xmi = min(x[i], xmi);
		xma = max(x[i], xma);
		ymi = min(y[i], ymi);
		yma = max(y[i], yma);
	}
	;
	sort(x + 1, x + n + 1);
	sort(y + 1, y + n + 1);
	for (int i = 1; i <= n; i++) {
		x[i] -= xmi;
		y[i] -= ymi;
	}
	/*
	for (int i = 1; i <= n; i++) {
		cout<<x[i]<<"  ";
	
	}
	cout<<endl;
		for (int i = 1; i <= n; i++) {
			cout<<y[i]<<"  ";
		
		}
	*/
	int ttx = 0, tty = 0;
	for (int i = 1; i <= n; i++) {

		ttx += x[i];


		tty += y[i];
	}
//cout<<ttx<<" "<<tty;
	xzhi[xmi + 2000000] = ttx;
	yzhi[ymi + 2000000] = tty;
//	cout<<ttx<<" "<<tty<<endl;
	if (ttx <= D)xtong[ttx]++;
	if (tty <= D)ytong[tty]++;
	for (int i = 1; i <= n; i++) {
		x[i] += xmi;
		y[i] += ymi;
	}

	for (int i = xmi - 1 + 2000000; i >= 0; i--) {
		xzhi[i] =	xzhi[i + 1] + n;
		if (xzhi[i] <= D)xtong[xzhi[i]]++;
		else break;

//		cout<<xzhi[i]<<endl;;
	}
	for (int i = ymi - 1 + 2000000; i >= 0; i--) {
		yzhi[i] =	yzhi[i + 1] + n;
		if (yzhi[i] <= D)ytong[yzhi[i]]++;
		else break;
		//	cout<<	yzhi[i]<<endl ;
	}
	int l = 	xj[xmi + 2000000], r = n - xj[xmi + 2000000  ];
	for (int i = xmi + 1 + 2000000; i <= 4000000; i++) {
		xzhi[i] =	xzhi[i - 1] + l - r;
		l +=	xj[i];
		r -= xj[i];

		if (xzhi[i] <= D)xtong[xzhi[i]]++;
	      if(i>xma&&xzhi[i]>=D)break;
	
	}

	l = 	yj[ymi + 2000000], r = n - yj[ymi + 2000000];
	for (int i = ymi + 1 + 2000000; i <= 4000000; i++) {
		yzhi[i] =	yzhi[i - 1] + l - r;
		l +=	yj[i];
		r -= yj[i];
		if (yzhi[i] <= D)ytong[yzhi[i]]++;
      if(i>yma+2000000&&yzhi[i]>=D)break;

	}

	xtongqzh[0] = xtong[0];
	ytongqzh[0] = ytong[0];

	for (int i = 1; i <=D; i++) {
		xtongqzh[i] = xtongqzh[i - 1] + xtong[i];

		ytongqzh[i]  = ytongqzh[i - 1] + ytong[i];
	}

/*	for (int i = 0; i <= D; i++) {
		cout << xtongqzh[i]<<" ";
	}
	cout<<endl;
	for (int i = 0; i <= D; i++) { 
		cout << xtongqzh[i]<<" ";
	}
	cout<<endl;*/
	int ttt = 0;
	for (int i = 0; i <= D; i++) {

		if (D - i >= 0)
			ttt += xtong[i] * ytongqzh[D - i] ;


	}
	cout << ttt;
}