1 条题解

  • 0
    @ 2025-10-10 15:45:26

    C :

    #include<stdio.h>
    void main()
    {
    	int n,i,j;
    	scanf("%d",&n);
    	//控制输出n行 
    	for(i=1;i<=n;i++)
    	{
    		//输出第i行空格的数量 
    		for(j=1;j<=n-i;j++)
    			printf(" ");
    		
    		//输出第i行*的数量
    		 for(j=1;j<=2*i-1;j++)
    			 printf("*");
    			 
    		 printf("\n");
    		 
    	}
    
    
    	
    	
    
    
    	
    	
    		
    }
    

    C++ :

    #include <iostream>
    using namespace std;
    
    int main(){ 
    	int i,j,n;
    	char c='A';
    	cin>>n;
    	for(i = 1;i <= n;i++){	
    		for(j = 1;j <= n - i;j++){
    			cout<<" ";
    		} 		
    		for(j = 1;j <= 2 * i - 1;j++){
    			cout<<"*";
    		} 	
    		cout<<endl;
    	} 	 
    }
    
    

    Pascal :

    var a,b,i,c,j,k:longint;
    begin
    read(a);
    b:=a-1;
    for i:=1 to a do
    begin
    for j:=b downto 1 do
    write(' ');
    b:=b-1;
    c:=i*2-1;
    for k:=1 to c do
    write('*');
    writeln;
    end;
    end.
    

    Java :

    import java.util.Scanner;
    public class Main{
    	public static void main(String[] args){
    		Scanner scanner=new Scanner(System.in);
    		int n=scanner.nextInt();
    		for(int i=1;i<=n;i++){
    			for(int j=1;j<=n-i;j++){
    				System.out.print(" ");
    			}
    			for(int j=1;j<=2*i-1;j++){
    				System.out.print("*");
    			}
    			System.out.println();
    		}
    	}
    }
    

    Python :

    n= int(input())
    for i in range (1, n + 1):
        for k in range (1, n - i + 1):
            print(' ', end='')
        for j in range (1, 2 * i):
            print('*', end = '')
        print();
     
    
    • 1

    信息

    ID
    67
    时间
    1000ms
    内存
    512MiB
    难度
    10
    标签
    递交数
    1
    已通过
    1
    上传者