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
Source
BestCoder Round #28
dfs 搜索即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
ll cal(ll x) {
ll ans = 0;
while (x) {
ans += (x % 10);
x /= 10;
}
return ans;
}
ll n;
int T;
int f[50];
bool fg;
int k;
int a[50];
bool dfs(int n, int stp) {
if (n == 1) {
fg = true;
return true;
}
if (fg)return true;
for (int i = stp; i < k; i++) {
if (n%a[i] == 0) {
if (dfs(n / a[i], i))return true;
}
}
return false;
}
int main()
{
//ios::sync_with_stdio(false);
rdint(T);
f[0] = 0; f[1] = 1;
for (int i = 2; i <= 45; i++) {
f[i] = f[i - 1] + f[i - 2];
}
while (T--) {
rdllt(n); fg = false;
ms(a); k = 0;
if (n == 0) {
cout << "Yes" << endl; continue;
}
for (int i = 3; i <= 45; i++) {
if (n%f[i] == 0) {
a[k] = f[i]; k++;
}
}
if (dfs(n, 0))cout << "Yes" << endl;
else cout << "No" << endl;
}
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
ll cal(ll x) {
ll ans = 0;
while (x) {
ans += (x % 10);
x /= 10;
}
return ans;
}
ll n;
int T;
int f[50];
bool fg;
int k;
int a[50];
bool dfs(int n, int stp) {
if (n == 1) {
fg = true;
return true;
}
if (fg)return true;
for (int i = stp; i < k; i++) {
if (n%a[i] == 0) {
if (dfs(n / a[i], i))return true;
}
}
return false;
}
int main()
{
//ios::sync_with_stdio(false);
rdint(T);
f[0] = 0; f[1] = 1;
for (int i = 2; i <= 45; i++) {
f[i] = f[i - 1] + f[i - 2];
}
while (T--) {
rdllt(n); fg = false;
ms(a); k = 0;
if (n == 0) {
cout << "Yes" << endl; continue;
}
for (int i = 3; i <= 45; i++) {
if (n%f[i] == 0) {
a[k] = f[i]; k++;
}
}
if (dfs(n, 0))cout << "Yes" << endl;
else cout << "No" << endl;
}
}
