【DFS】hdu 5167 Fibonacci

Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2861    Accepted Submission(s): 744

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 numberT 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

3417233

 

Sample Output

YesNoYes

题意：一个数能否分解为斐波那契数的乘积

思路：先暴力打斐波那契表，然后对每个数dfs

/// AC代码

#include <iostream>#include <cstdio>#include <string.h>#include <math.h>#include <stdlib.h>#include <algorithm>using namespace std;#define mx 1000000000int a[50];void get (){    a[0] = 0;    a[1] = 1;    for (int i = 2; a[i - 1] < mx; i++)    {        a[i] = a[i - 1] + a[i - 2];    }}int dfs(int n, int pos){    if (n == 1)    {        return 1;    }    for (int i = pos; i >= 3; i--)    {        if (n % a[i] == 0)        {            if (dfs(n / a[i], i) == 1)            {                return 1;            }        }    }    return 0;}int main(){    get();    int n, t;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        if (n == 0)        {            printf("Yes\n");            continue;        }        if (dfs(n, 45) == 1)        {            printf("Yes\n");        }        else        {            printf("No\n");        }    }    return 0;}