bestcoder#28 1002 dfs

bestcoder#28 1002  dfs

Fibonacci

Accepts: 40

Submissions: 996

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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

题意：判断n是否为一个或多个fib数的乘积
思路：dfs，从大数往小数遍历更省时

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<cstring>

#include<algorithm>

#include<vector>



using namespace std;



const int maxn=1000100;

const int INF=1000000000;



int T;

int n;

vector<int> fib;



inline int read()

{

    char c=getchar();

    while(c!='-'&&!isdigit(c)) c=getchar();

    int f=0,tag=1;

    if(c=='-'){

        tag=-1;

        f=getchar()-'0';

    }

    else f=c-'0';

    while(isdigit(c=getchar())) f=f*10+c-'0';

    return f*tag;

}



void creat_fib()

{

    fib.clear();

    fib.push_back(0);

    fib.push_back(1);

    int x=fib[0]+fib[1];

    while(x<=INF&&x>=0){

        fib.push_back(x);

        x=fib[fib.size()-1]+fib[fib.size()-2];

    }

}



bool dfs(int n,int t)

{

    if(t==2) return false;

    while(true){

        if(dfs(n,t-1)) return true;

        if(n%fib[t]) break;

        n/=fib[t];

    }

    if(n==1) return true;

    return false;

}



int main()

{

    cin>>T;

    creat_fib();

    while(T--){

        n=read();

        if(n==0||n==1||dfs(n,fib.size()-1)) printf("Yes\n");

        else printf("No\n");

    }

    return 0;

}

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