F - Friends

Mike has many friends. Here are nine of them: Alice, Bob, Carol, Dave, Eve, Frank, Gloria, Henry and Irene. 

Mike is so skillful that he can master nn languages (aka. programming languages). 

His nine friends are all weaker than he. The sets they can master are all subsets of Mike's languages. 

But the relations between the nine friends is very complex. Here are some clues. 

1. Alice is a nice girl, so her subset is a superset of Bob's. 
2. Bob is a naughty boy, so his subset is a superset of Carol's. 
3. Dave is a handsome boy, so his subset is a superset of Eve's. 
4. Eve is an evil girl, so her subset is a superset of Frank's. 
5. Gloria is a cute girl, so her subset is a superset of Henry's. 
6. Henry is a tall boy, so his subset is a superset of Irene's. 
7. Alice is a nice girl, so her subset is a superset of Eve's. 
8. Eve is an evil girl, so her subset is a superset of Carol's. 
9. Dave is a handsome boy, so his subset is a superset of Gloria's. 
10. Gloria is a cute girl, so her subset is a superset of Frank's. 
11. Gloria is a cute girl, so her subset is a superset of Bob's. 

Now Mike wants to know, how many situations there might be. 

Input

The first line contains an integer TT(T≤20T≤20) denoting the number of test cases. 

For each test case, the first line contains an integer nn(0≤n≤30000≤n≤3000), denoting the number of languages. 

Output

For each test case, output ''Case #t:'' to represent this is the t-th case. And then output the answer. 

Sample Input

2
0
2

Sample Output

Case #1: 1
Case #2: 1024

Hint

题意：

其实就是让你求32^n,注意3000会爆 int long long 所以需要使用大数运算！

 思路：
此题凭借运气蒙对了，，，其实就是32的n次方，队友的提示使我尝试了一下32  结果 ac

代码：

#include <bits/stdc++.h>
using namespace std;
const int s=100;
int len;
int num[50001];
int main()
{
	int T,ca=0;
	scanf("%d",&T);
	while(T--)
	{
		ca++;
		int n;
		scanf("%d",&n);
		memset(num,0,sizeof(num));
		num[1]=1;
		len=1;
		for(int i=1;i<=n;i++)
		{
			int c=0;
			for(int j=1;j<=len;j++)
			{
				num[j]*=32;
				num[j]+=c;
				c=num[j]/s;
				num[j]%=s;
			}
			while(c)
			{
				num[++len]=c%s;
				c/=s;
			}
		
		}
		cout<<"Case #"<<ca<<":"<<" ";
		for(int i=len;i>0;i--)
		{
			if(i!=len)
			printf("%02d",num[i]);
			else 
			printf("%d",num[i]);
		}
		cout<<endl;
	}
	return 0;
}