View Code of Problem 53

#include<stdio.h>
void main()
{int i,j,n,k,m;
scanf("%d",&n);
for (i=1;i<=n;i++)//打印上半部分
{
	for(j=1;j<=n-i;j++)//打印空格,个数递减
	{printf(" ");}
	for(k=1;k<=2*i-1;k++)//打印* ,个数为2k-1
	printf("*");
	printf("\n");
}
	
for(i=1;i<=n-1;i++)//打印下半部分
{
	for(j=1;j<=i;j++)//打印空格
	{printf(" ");}
for (m=1;m<=2*(n-i)-1;m++)//找一下几何关系即可得到2*(n-1)-1这个关系
{ printf("*");}
printf("\n");
}
}

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