【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）：

 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 }

View Code