链接地址：http://acm.hdu.edu.cn/showproblem.php?pid=5167
Total Submission(s): 474 Accepted Submission(s): 126
Problem Description
Following is the recursive definition of Fibonacci sequence:
Now we need to check whether a number can be expressed as the product of numbers in the 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
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;
}
