HDU 5167 Fibonacci（BestCoder Round #28）

Problem Description：

Following is the recursive definition of Fibonacci sequence:

Fi=⎧⎩⎨01Fi−1+Fi−2i = 0i = 1i > 1

Now we need to check whether a number can be expressed as the product of numbers in the Fibonacci sequence.

 

Input：

There is a number T shows there are T test cases below. (T≤100,000)
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

题意：判断一个数n是不是Fibonacci序列中数的乘积（可以是同一个数）。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9;

typedef long long LL;

LL n, a[60];
int k, flag;

void Solve() ///打表，并找到满足题意的最大的Fibonacci数的下标k
{
    int i;

    a[0] = 0; a[1] = 1;

    for (i = 2; i <= 50; i++)
    {
        a[i] = a[i-1]+a[i-2];

        if (a[i] >= MOD)
        {
            if (a[i] > MOD) i--;
            k = i;
            break;
        }
    }
}

void DFS(int num, LL m)
{
    int i;

    if (m == 1) flag = 1;

    if (flag == 1)
        return ;

    if (num < 3) return ; 

    if (m % a[num] == 0)
        DFS(num, m/a[num]);

    DFS(num-1, m);
}

int main ()
{
    int T;

    Solve(); 

    scanf("%d", &T);

    while (T--)
    {
        scanf("%lld", &n);

        flag = 0;

        if (n == 0)
        {
            printf("Yes
");
            continue;
        }

        DFS(k, n); 

        if (flag == 1) printf("Yes
");
        else printf("No
");
    }

    return 0;
}