【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 = 100;
int f[maxn], tmp[maxn], j, k;
void Fibonacci()
{
    f[0] = 0; f[1] = 1;
    k = 2;
    while(f[k-1] + f[k-2] <= 1000000000)
    {
        f[k] = f[k-1] + f[k-2];
        k++;
    }
}
bool Judge(int x, int step)//注意这里没有step的话会tle。。。
{
    if(x == 1) return true;
    for(int i = step; i < j; i++)
    {
        if(x%tmp[i] == 0)
        {
            if(Judge(x/tmp[i], i))
                return true;
        }
    }
    return false;
}
int main()
{
    Fibonacci();
    int T;
    scanf("%d", &T);
    for(int i = 0; i < T; i++)
    {
        int n;
        scanf("%d", &n);
        if(!n) printf("Yes
");
        else
        {
            j = 0;
            for(int i = 3; i < k; i++)
            {
                if(n%f[i] == 0)
                    tmp[j++] = f[i];
            }
            if(Judge(n, 0)) printf("Yes
");
            else printf("No
");
        }
    }
    return 0;
}