【Fibonacci】BestCoder #28B Fibonacci

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）：

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;
 const int maxn = ;
 int f[maxn], tmp[maxn], j, k;
 void Fibonacci()
 {
     f[] = ; f[] = ;
     k = ;
     while(f[k-] + f[k-] <= )
     {
         f[k] = f[k-] + f[k-];
         k++;
     }
 }
 bool Judge(int x, int step)//注意这里没有step的话会tle。。。
 {
     if(x == ) return true;
     for(int i = step; i < j; i++)
     {
         if(x%tmp[i] == )
         {
             if(Judge(x/tmp[i], i))
                 return true;
         }
     }
     return false;
 }
 int main()
 {
     Fibonacci();
     int T;
     scanf("%d", &T);
     for(int i = ; i < T; i++)
     {
         int n;
         scanf("%d", &n);
         if(!n) printf("Yes\n");
         else
         {
             j = ;
             for(int i = ; i < k; i++)
             {
                 if(n%f[i] == )
                     tmp[j++] = f[i];
             }
             if(Judge(n, )) printf("Yes\n");
             else printf("No\n");
         }
     }
     return ;
 }