#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n / 2; i++)
    {
        for (int j = 0; j < i; j++)
        {
            cout << " ";
        }

        cout << "*";

        for (int j = n / 2 - 1; j > i; j--)
        {
            cout << " ";
        }

        cout << "*";

        for (int j = n / 2 - 1; j > i; j--)
        {
            cout << " ";
        }

        cout << "*";

        cout << endl;
    }
    
    for (int i = 0; i < n; i++)
    {
        cout << "*";
    }
    cout << endl;

    for (int i = n / 2 - 1; i >= 0 ; i--)
    {
        for (int j = 0; j < i; j++)
        {
            cout << " ";
        }

        cout << "*";

        for (int j = 0; j < n / 2 - 1 - i; j++)
        {
            cout << " ";
        }

        cout << "*";

        for (int j = 0; j < n / 2 - 1 - i; j++)
        {
            cout << " ";
        }
        
        cout << "*";

        cout << endl;
    }
    return 0;
}

#include<bits/stdc++.h> using namespace std; int main() { int n;

cin>>n;

int mid=n/2+1;

int a[105][105]={0};

for(int i=1; i<=n; i++) //i代表x 
{
	for(int j=1; j<=n; j++) //j代表y 坐标(i,j) 
	{
		if(i==j)
			a[i][j]=1;
		else if(i==mid||j==mid)
			a[i][j]=1;
		else if(i+j==n+1)
			a[i][j]=1; 
	}
}

for(int i=1; i<=n; i++)
{
	for(int j=1; j<=n; j++)
	{
		if(a[i][j]==1)
			cout<<"*";
		else
			cout<<" ";
	}
	cout<<endl; //每行结束回撤 
}


return 0;

} /* 0 1 2 3 4 5 6 y(j)

1 *

2

3

4

5

6

x(i) */

注意到米字需要输出两条对角线，一个横列和一个纵列的*，自然联想到使用二维数组来做这道题。

该有可能因为分给的太少的原因，会变成标红的ac

#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
#include<math.h>

int main(){
	int n,x;
    scanf("%d", &n);
    x=(n-1)/2;
    char a[n][n];
    memset(a,' ',sizeof(a));
    int i,j;
    for(i=0;i<n;i++){
    	j=x;
    	a[i][j]='*';
	}
	for(j=0;j<n;j++){
		i=x;
		a[i][j]='*';
	}
	for(i=0;i<n;i++){
		j=i;
			a[i][j]='*';
		
	}
	for(i=0;i<n;i++){
		j=n-1-i;
			a[i][j]='*';
		
	}	
	for(i=0;i<n;i++){
		for(j=0;j<n;j++){
			printf("%c",a[i][j]);
		}
		printf("\n");
	}
    return 0;
}