1 条题解
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0
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