hdu 5167 Fibonacci(预处理)

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：判断一个数能否由任意个菲波那媞数相乘得到，能的话输出“Yes”，否则输出“No”

思路：首先要处理菲波那媞数组，然后用一个队列来实现菲波那媞数的相乘，用一个map数组来标记。具体看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
#define N 46
#define ll long long
ll f[N];
map<ll,bool>mp;
void init()
{
    mp.clear();
    queue<ll>q;
    f[0]=0;
    f[1]=1;
    mp[0]=true;
    mp[1]=true;
    q.push(0);
    q.push(1);
    for(int i=2;i<N;i++)
    {
        f[i]=f[i-1]+f[i-2];
        mp[f[i]]=true;
        q.push(f[i]);
    }
    while(!q.empty())
    {
        ll tmp=q.front();
        q.pop();
        for(int i=0;i<N;i++)
        {
            ll cnt=tmp*f[i];
            if(cnt>1000000000L)
              break;
            if(mp[cnt]) continue;
            mp[cnt]=true;
            q.push(cnt);
        }
    }
    
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n;
        scanf("%I64d",&n);
        if(mp[n]==true)
          printf("Yes
");
        else 
          printf("No
");
        
    }
    return 0;
}