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[]=;
     f[]=;
     mp[]=true;
     mp[]=true;
     q.push();
     q.push();
     for(int i=;i<N;i++)
     {
         f[i]=f[i-]+f[i-];
         mp[f[i]]=true;
         q.push(f[i]);
     }
     while(!q.empty())
     {
         ll tmp=q.front();
         q.pop();
         for(int i=;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\n");
         else
           printf("No\n");

     }
     return ;
 }