链接地址： http://acm.hdu.edu.cn/showproblem.php?pid=5167

Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K
(Java/Others)
Total Submission(s): 474 Accepted Submission(s): 126

Problem Description

Following is the recursive definition of Fibonacci sequence:

F i = ⎧ ⎩ ⎨ 0 1 F i − 1 + F i − 2 i = 0 i = 1 i > 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

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include <sstream>
#include <time.h>
#include <utility>
#include <malloc.h>

using namespace std;

__int64 p[100];
int T;
__int64 n;

int main()
{
    p[0] = 0; p[1] = 1;
    for (int i = 2; i <= 46; i++)
        p[i] = p[i - 1] + p[i - 2];

    set<__int64> q;
    set<long long> ::iterator it;
    q.clear();
    q.insert(0);
    q.insert(1);
    for(it=q.begin();it!=q.end();it++)
    {
         for(int j=3;j<=45;j++)
        {
            int tt = *it;
            __int64 tmp = tt*p[j];
            if (tmp <= 1000000000)
                q.insert(tmp);
        }
    }
    cin>>T;
    while (T--)
    {
        int ok=0;
        cin>>n;
        if (q.find(n) != q.end())
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}