Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2391    Accepted Submission(s): 609


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 題意：問n是否能由多個斐波那契數相乘而來

思路：直接爆搜會TLE，所以這裡有個剪枝技巧，我們從大小大列舉，這一次搜尋選擇了i,那麼我們下次只需要列舉3~i即可，相等於從序列最大的數開始向下列舉。

注意0,1,2不能算進去，0,和1要特判，不然會死迴圈。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
#define N 100000
long long f[N];
int flag,cnt;
long long n;
void init()
{
    f[0]=0;
    f[1]=1;
    cnt=2;
    for(; f[cnt-1]<=1000000000; cnt++)
        f[cnt]=f[cnt-1]+f[cnt-2];
}
void dfs(long long k,int num)
{
    if(k==1){
        flag=1;
        return;
    }
    for(int i=num;i>=3;i--)
        {
            if(f[i]>k) continue;
            if(k%f[i]==0)
            dfs(k/f[i],i);
            if(flag) return;
        }
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&n);
        if(!n||n==1) printf("Yes\n");
        else
        {
            flag=0;
            dfs(n,cnt-1);
            if(flag) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}