hdu 5167 Fibonacci 打表

Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

 

Source

BestCoder Round #28

题意：问一个数n，能否由斐波那契数列中的某些数乘积组成；

思路：打表，能组成的全弄出来；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
map<ll,int>m;
ll f[N];
priority_queue<ll,vector<ll>,greater<ll> >q;
void init()
{
    int pre=0;
    int now=1;
    f[0]=0;
    f[1]=1;
    for(int i=2;i<=44;i++)
        f[i]=f[i-1]+f[i-2];
    for(int i=0;i<=44;i++)
        if(!m[f[i]])
        q.push(f[i]),m[f[i]]=1;
    while(!q.empty())
    {
        ll v=q.top();
        q.pop();
        for(int i=1;i<=44;i++)
        {
            if(f[i]*v<inf&&!m[f[i]*v])
            {
                m[f[i]*v]=1;
                q.push(f[i]*v);
            }
        }
    }
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        if(m[n])
            printf("Yes
");
        else
            printf("No
");
    }
    return 0;
}