HDU_5167_Fibonacci

题意：判断一个数（0≤n≤1,000,000,000）能否表示成菲波数的乘积（即该数的所有因子都是菲波数）；
思路：先把0≤n≤1,000,000,000内的菲波数放到一个数组中，再在对该数的分解时dfs。

#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <list>
#include <stack>
#include <deque>
#include <queue>
#include <vector>
#include <algorithm>
#include <functional>

#define PI acos(-1.0)
#define eps 1e-10
#define INF 0x7fffffff
#define debug(x) cout << "--------------> " << x << endl

typedef long long LL;
typedef unsigned long long ULL;

using namespace std;

long long fibo[50];
int mark[100001] = {0};
bool dfs(LL num, int k)
{
if (num < 100001 && mark[num] != 0)
{
return mark[num] - 1;
}
if (num == 1)
{
return 1;
}
for (int i = k; i > 2; --i)
if (num % fibo[i] == 0 && dfs(num / fibo[i], i))
{
if (num < 100001)
{
mark[num] = 2;
}
return 1;
}

if (num < 100001)
{
mark[num] = 1;
}
return 0;
}

int main()
{
fibo[0] = 0;
fibo[1] = 1;
for (int i = 2; i < 50; ++i)
{
fibo[i] = fibo[i - 1] + fibo[i - 2];
}
int T;
scanf("%d", &T);
LL num;
while (T--)
{
scanf("%lld", &num);
if (num == 0 || num == 1)
{
printf("Yes\n");
}
else
{
dfs(num, 45) ? printf("Yes\n") ：printf("No\n");
}
}

return 0;
}