题解

用map储存，再遍历时，加上$a_i-m$的数量即可
注意，没说先后顺序，不用除2

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
map <int, int> mp;
int a[200005];
long long ct;
int read(){
	char c = getchar();
	int x = 0;
	while(!isdigit(c)) c = getchar();
	while(isdigit(c)){
		x = (x << 3) + (x << 1) + c - '0';
		c = getchar();
	}return x;
}
int main(){
	int n = read(), m = read();
	long long ct = 0;
	for(int i = 1; i <= n; i++){
		a[i] = read();
		mp[a[i]]++;//记录
	}for(int i = 1; i <= n; i++){
		if(a[i] >= m) ct += mp[a[i] - m];//加上
	}
	cout << ct;

	return 0;
}

查询$mp[i]$时间复杂度:$O(\log_2 a[i])$
时间复杂度:$O(n\log_2 a[i])$