hdu-5167 Fibonacci

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:

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

3417233

 

Sample Output

YesNoYes

 

#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;}