View Code of Problem 53

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
	int n=0;
	scanf("%d",&n);
	int k=0,i=0;
	for(k=i-n;k<=n-1;k++)
	{
		i=n-abs(k);
		//先打印空格
		//abs(K)=4,空格为4个，故长度与abs(k)等长
		for(int j=1;j<=abs(k);j++)
		{
			printf(" ");
		}
		//打印星号
		//行数*2-1
		for(int j=1;j<=2*i-1;j++)
		{
			printf("*");
		}
		printf("\n");
	}
	printf("\n");
}