HDU 5167 Fibonacci（DFS暴力搜索）

Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1905    Accepted Submission(s): 479

Problem Description

Following is the recursive definition of Fibonacci sequence:

斐波那契数列的递归定义如下：

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.

现在我们需要判断一个数是否能表示为斐波那契数列中的数的乘积。

 

Input

There is a number T shows there are T test cases below. (T≤100,000)

有多组数据，第一行为数据组数$T$（$T\le 100,000$）。

For each test case , the first line contains a integers n , which means the number need to be checked. 

对于每组数据有一个整数$n$，表示要判断的数字。

0≤n≤1,000,000,000

 

Output

For each case output "Yes" or "No".

对于每组数据，如果可以输出“Yes”，否则输出"No"。

 

Sample Input

3417233

 

Sample Output

YesNoYes

 

Source

BestCoder Round #28

 

解题思路：由于斐波那契数列变换比较快，在1000000000内的斐波那契数只有不到50，可以对每一个n搜索n的斐波那契因数，如果找得到斐波那契因数，输出Yes，否则输出No。

代码如下：

#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <algorithm>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#include <ctime>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;ll f[55];bool DFS(ll n,ll m){    if(n<=6)        return true;    for(ll i=3; i<=m; i++)    {        if(n%f[i])            continue;        if(DFS(n/f[i],i))            return true;    }    return false;}int main(){    ll i,j,k,n,t;    f[0]=0ll;    f[1]=1ll;    for(i=2; i<=45; i++)        f[i]=f[i-1]+f[i-2];    /*    for(i=0;i<=50;i++)        printf("%I64d\n",f[i]);    */    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d",&n);        if(DFS(n,45))            printf("Yes\n");        else            printf("No\n");    }    return 0;}