【CF ＃277.5 (Div. 2)】B. BerSU Ball

Through the tunnel

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The question , Find the elements in two arrays , Is there an absolute value that does not exceed 1.（ If exist , Then form a pair , And cannot be paired with other elements ）
Find out how many pairs can be matched .

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Ideas ： The ability to represent the dancing skills of boys and girls , Sort .
Due to the small data range , So we used $O(n^2)$ Time complexity of
You can also use the double pointer method for this problem , Let's do some optimization .

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$O(n^2)$ Methods , Double cycle

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int n,t,m;
int a[N],b[N],c[N];
bool vis[N];
int ans;
int main(){
    
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	sort(a,a + n);
	cin >> m;
	for(int j = 0; j < m; j++) cin >> b[j]; 
	sort(b, b + m);
	for(int j = 0; j < m; j++){
    
		for(int i = 0 ; i < n; i++){
    
			if(abs(a[i] - b[j]) <= 1 && !vis[i]){
    
				ans++;
				vis[i] = 1;
				break;
			}
		}	
	}
	
	cout << ans << endl;
	return 0;
}

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Double finger needling
Ideas ：
Sort two arrays
Two pointers point to the first element of two arrays respectively .
If the absolute value of the element that these two pointers currently point to is less than or equal to 1, be ans++,
otherwise , Compare the size of two pointers pointing to elements , To determine which pointer moves toward the end of the array

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int n,t,m;
int a[N],b[N],c[N];
bool vis[N];
int ans;
int main(){
    
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	sort(a,a + n);
	cin >> m;
	for(int j = 0; j < m; j++) cin >> b[j]; 
	sort(b, b + m);
	for(int i = 0, j = 0; i < n && j < m;){
    
		if(abs(a[i]-b[j]) <= 1){
    
			i++;
			j++;
			ans++;
		}
		else if(a[i] < b[j]){
    
			i++;
		}
		else if(a[i] > b[j]){
    
			j++;
		}
	}
	
	cout << ans << endl;
	return 0;
}