陈老师的多校联合 20140809 C题

http://vjudge.net/contest/view.action?cid=51408#problem/C

Description





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 Nsides
of a brick cannot be anything other than fibonacci numbers. A fibonacci sequence is given below:

[align=center]1, 2, 3, 5, 8, 13, 21...[/align]



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 containingN(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 Casex: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

题目大意：n维几何体，判断至少用多少n维且各维边长都为Fibonacci数的砖。

解题思路：求出n个数每个数至少是几个个Fibonacci数（可以重复）的和，然后这样把n个数的统计数乘起来就是答案。

#include <iostream>
#include <cstring>
#include <set>
#include <cstdio>
using namespace std;

typedef long long LL;

const int maxn =55;

LL f[maxn],x[20];

set<LL>fib;
set<LL>::iterator it;
void init()
{
f[0]=1;
f[1]=1;
fib.insert(1);
for(int i=2; i<maxn; i++)
{
f[i]=f[i-1]+f[i-2];
fib.insert(f[i]);
}
}

int main()
{
int n,cas=1,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%lld",&x[i]);
init();
int i,j;
long long ans=1;
int cnt=1;
for(i=0; i<n; i++)
{
//每个x[i]至少由几个斐波那契数组成（可以重复利用一个数多次）
if(fib.find(x[i])==fib.end())
{
cnt=1;
int s=x[i];
while(fib.find(s)==fib.end())
{
int t=lower_bound(f,f+55,s)-f;
s=s-f[t-1];
cnt++;
}
ans*=cnt;
}
}
printf("Case %d: %lld\n",cas++,ans);
}
return 0;
}