UVA 4855 Hyper Box

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
",cas++,sum);
	}
	return 0;
}