Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 795    Accepted Submission(s): 213

Problem Description

Following is the recursive definition of Fibonacci sequence:

Fi=???01Fi−1+Fi−2i = 0i = 1i > 1

Fi=　　0, 　　　　　 i = 0　　1, 　　　　　 i = 1　　Fi−1+Fi−2, 　i > 1

Now we need to check whether a number can be expressed as the product of numbers in the Fibonacci sequence.(一开始没理解题意，其实我觉得这题意也不大好=。 =，题解说是Fib乘积，但和为什么不行呢，路过的可以帮忙解答一下)

Input

There is a number T shows there are T test cases below. ( T < 100000 )
For each test case , the first line contains a integers n , which means the number need to be checked.
0≤n≤1,000,000,000

Output

For each case output "Yes" or "No".

Sample Input

3
4
17
233

Sample Output

Yes
No
Yes

Note：

1、Fibonacci序列在1~1000000000范围内只有43个数。

2、判断一个数是几个数乘积的递归算法（掌握）；

代码（看过题解）：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 using namespace std;
 6 const int maxn = 100;
 7 int f[maxn], tmp[maxn], j, k;
 8 void Fibonacci()
 9 {
10     f[0] = 0; f[1] = 1;
11     k = 2;
12     while(f[k-1] + f[k-2] <= 1000000000)
13     {
14         f[k] = f[k-1] + f[k-2];
15         k++;
16     }
17 }
18 bool Judge(int x, int step)//注意这里没有step的话会tle。。。
19 {
20     if(x == 1) return true;
21     for(int i = step; i < j; i++)
22     {
23         if(x%tmp[i] == 0)
24         {
25             if(Judge(x/tmp[i], i))
26                 return true;
27         }
28     }
29     return false;
30 }
31 int main()
32 {
33     Fibonacci();
34     int T;
35     scanf("%d", &T);
36     for(int i = 0; i < T; i++)
37     {
38         int n;
39         scanf("%d", &n);
40         if(!n) printf("Yes\n");
41         else
42         {
43             j = 0;
44             for(int i = 3; i < k; i++)
45             {
46                 if(n%f[i] == 0)
47                     tmp[j++] = f[i];
48             }
49             if(Judge(n, 0)) printf("Yes\n");
50             else printf("No\n");
51         }
52     }
53     return 0;
54 }