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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<stdlib.h>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<map>
 9 using namespace std;
10 #define N 46
11 #define ll long long
12 ll f[N];
13 map<ll,bool>mp;
14 void init()
15 {
16     mp.clear();
17     queue<ll>q;
18     f[0]=0;
19     f[1]=1;
20     mp[0]=true;
21     mp[1]=true;
22     q.push(0);
23     q.push(1);
24     for(int i=2;i<N;i++)
25     {
26         f[i]=f[i-1]+f[i-2];
27         mp[f[i]]=true;
28         q.push(f[i]);
29     }
30     while(!q.empty())
31     {
32         ll tmp=q.front();
33         q.pop();
34         for(int i=0;i<N;i++)
35         {
36             ll cnt=tmp*f[i];
37             if(cnt>1000000000L)
38               break;
39             if(mp[cnt]) continue;
40             mp[cnt]=true;
41             q.push(cnt);
42         }
43     }
44
45 }
46 int main()
47 {
48     init();
49     int t;
50     scanf("%d",&t);
51     while(t--)
52     {
53         ll n;
54         scanf("%I64d",&n);
55         if(mp[n]==true)
56           printf("Yes\n");
57         else
58           printf("No\n");
59
60     }
61     return 0;
62 }