# include <bits/stdc++.h>

using namespace std;

int main(){
	int n;
	cin>>n;
	for(int i = 1;i<=n;i++){
		for(int j = 1;j<=n-i;j++)cout<<" ";
		for(int j = 0;j<i;j++)cout<<(char)(65+j);
		for(int j = i-2;j>=0;j--)cout<<(char)(65+j);
		cout<<"\n";
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int n,c=65;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++){
	for(int j=1;j<=n-i;j++)cout<<" ";
	for(int j=1;j<=i;j++)cout<<(char)c++;
	c-=2;
	for(int j=1;j<=i-1;j++)cout<<(char)c--;
	c = 65;
	cout<<endl;
	}
	return 0;
}

这题跟#958图形处理问题七一模一样呀

#include<queue>
#include<math.h>
#include<stack>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<iomanip>
#include<string.h>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N=1e6+10;
const int INF=0x3f3f3f3f;
int main()
{
	int n;
    cin >> n;
    for (int i=1;i<=n;i++)
	{
        for (int j=1;j<=n-i;j++)
		{
            cout<<" ";
        }
        for (int j=1;j<=i;j++)
		{
            cout<<char('A'-1+j);
        }
        for (int j=i-1;j>0;j--)
		{
        	cout<<char('A'-1+j);
		}
        cout<<endl;
    }
	return 0;
}