题目描述

源代码

#include<iostream>
using namespace std;
int a[1005];
int main()
{
    
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
    
		scanf("%d", &a[i]);
	}
	int i, j;
	int count = 0;
	for (i = 0; i < n; i++)
	{
    
		for (j = i + 1; j < n; j++)
		{
    
			if (abs(a[i] - a[j]) == 1)
			{
    
				count++;
			}
		}
	}
	cout << count << endl;

	return 0;
}

关于这题

Using violence here The number of comparing two poor If the absolute value of1 符合条件 count++;