Luke'family

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Luke's family

TimeLimit: 1 Second MemoryLimit: 32Megabyte

Totalsubmit: 80 Accepted: 22

Description

In the stars’ family, Luke only has seen his father andgrandfather. He really wants to know what his ancestors look like.We numbered Luke as the No.0 star. Luke’s father is No.1 star, andthe father of Luke’s father is No.2 star, and so on.
Luke looks like：
*
Luke’s father looks like：

*
* * * *
* * *
* * * *
*

The father of Luke’s father looks like：

*
**
* * * * * * *
* * * * * *
* * * * *
* * * * * *
* * * * * * *
**
*

Input

There are multiple test cases.
In each test case, the first line is an integer n(1<=n<=90).

Output

For each test case, first output a line “Case T:” T starts with 1.And then output what the No.n star looks like.

Sample Input

0
1
2
3

Sample Output

Case 1:
*
Case 2:
*
* * * *
* * *
* * * *
*
Case 3:
*
**
* * * * * * *
* * * * * *
* * * * *
* * * * * *
* * * * * * *
**
*
Case 4:
   *
   * *
  * * *
* * * * * * * * * *
* * * * * * * * *
* * * * * * * *
* * * * * * *
* * * * * * * *
* * * * * * * * *
* * * * * * * * * *
  * * *
   * *
   *

Source

Luke

#include<iostream>#include<stdio.h>#include<algorithm>#include<string>#include<string.h>#include<stdlib.h>#include<math.h>using namespacestd;int main(){    int n,i,j,k,s,t,T=1;    while(~scanf("%d",&n)){      printf("Case%d:\n",T++);      if(n==0)           printf("*\n");      else {          s=3*n;          for(i=0;i<n;i++){              s--;             for (j=0;j<s; j++)                printf("");             for (k=0; k<2*i+2;k++) {                if(k%2)                    printf("*");                else                     printf(" ");              }             printf("\n");           }          t=-1;          s=3*n+1;          for (i=0; i<n+1; i++) {              t++;             for (k=0; k<t; k++)                printf("");             for (j=i; j<s; j++){                 if(j==i)                    printf("*");                else                     printf(" *");              }             printf("\n");           }           t=n;          s=2*n+1;          for (i=0; i<n; i++) {              t--;              s++;             for (k=0; k<t; k++) {                printf("");              }             for (j=0; j<s; j++) {                if(j==0)                    printf("*");                else                     printf(" *");              }             printf("\n");           }          s=2*n;          for(i=n;i>0;i--){              s++;             for (j=0;j<s-1; j++)                printf("");             for (k=2*i; k>0; k--) {                if(k%2)                    printf("*");                else                     printf(" ");              }             printf("\n");           }       }    }    return 0;}

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