1 条题解
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0
C :
int main(){ int n,i,j,z; scanf("%d",&n); //控制输出的行数 for(i = 1;i <= n;i++){ //控制每行输出空格量 for(j = 0;j < n - i;j++){ printf(" "); } //控住输出的* for(z=1;z <= 2 * i + 1;z++){ printf("*"); } //负责换行 printf("\n"); } }C++ :
#include<iostream> using namespace std; int main(){ int i,j,k,n; cin>>n; for(i = 1;i <= n;i++) { for(j = 1;j <= n - i;j++) { cout<<" "; } for(k = 1;k <= 2 * i + 1;k++) { cout<<"*"; } cout<<endl; } return 0; }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; write('**'); 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 x in range(1, 2 * i +2): print('*', end = '') print()
- 1
信息
- ID
- 68
- 时间
- 1000ms
- 内存
- 512MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者