You live in the universe X where all the

physical laws and constants are different

from ours. For example all of their objects

are N-dimensional. The living beings

of the universe X want to build an

N-dimensional monument. We can consider

this N dimensional monument as an

N-dimensional hyper-box, which can be

divided into some N dimensional hypercells.

The length of each of the sides of

a hyper-cell is one. They will use some

N-dimensional bricks (or hyper-bricks) to

build this monument. But the length of

each of the N sides of a brick cannot be

anything other than fibonacci numbers. A

fibonacci sequence is given below:

1, 2, 3, 5, 8, 13, 21, . . .

As you can see each value starting from 3 is the sum of previous 2 values. So for N = 3 they can

use bricks of sizes (2,5,3), (5,2,2) etc. but they cannot use bricks of size (1,2,4) because the length 4

is not a fibonacci number. Now given the length of each of the dimension of the monument determine

the minimum number of hyper-bricks required to build the monument. No two hyper-bricks should

intersect with each other or should not go out of the hyper-box region of the monument. Also none of

the hyper-cells of the monument should be empty.

Input

First line of the input file is an integer T (1 ≤ T ≤ 100) which denotes the number of test cases. Each

test case starts with a line containing N (1 ≤ N ≤ 15) that denotes the dimension of the monument

and the bricks. Next line contains N integers the length in each dimension. Each of these integers will

be between 1 and 2000000000 inclusive.

Output

For each test case output contains a line in the format Case x: M where x is the case number (starting

from 1) and M is the minimum number of hyper-bricks required to build the monument.

Sample Input

2

2

4 4

3

5 7 8

Sample Output

Case 1: 4

Case 2: 2

题意： 给一个n维空间的的物体，给出每一维的长度。问有最少几个比它体积小的物体组成它，要求这些物体的边必须是斐波那契数列

里边的数。

思路： 假设边长是斐波那契数就无论他，假设不是，比这个边长小的最大的斐波数减起，一直减到0。减了几个斐波数。也就是这条边

最少分解成几个斐波数，最后每一维相乘即为结果。

#include<stdio.h>

#include<string.h>

int fb[60];

int main(){

	int t,ok,n,cas=1;

	int a[20];

	fb[1]=1; fb[2]=2;

	for(int i=3;i<55;i++)

		fb[i]=fb[i-1]+fb[i-2];

	scanf("%d",&t);

	while(t--){

		int cnt=0;

		long long sum=1;//结果不用long long 会错

		scanf("%d",&n);

		for(int i=0;i<n;++i)

			scanf("%d",&a[i]);

		for(int i=0;i<n;++i){

			cnt=ok=0; int k;

			for(int j=1;j<55;j++){

				if(a[i]==fb[j]){

					ok=2;

					break;

				}

				if(a[i]<fb[j]){

					ok=1;

					k=j;

					break;

				}

			}

			if(ok==1){

				int x=a[i];

				while(x){

					while(fb[k]>x)

						k--;

					x-=fb[k];

					cnt++;

				}

			}

			if(ok!=2)//ok==2时证明这条边是斐波数

				sum*=cnt;//注意是相乘。，

		}

		printf("Case %d: %lld\n",cas++,sum);

	}

	return 0;

}