http://acm.hdu.edu.cn/showproblem.php?pid=5167
Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3388 Accepted Submission(s): 886
Problem Description
Following is the recursive definition of Fibonacci sequence:
Now we need to check whether a number can be expressed as the product of numbers in the 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
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
Recommend
#include <iostream>
using namespace std;
int a[100];
int ct, flag;
void init(){
a[0] = 0;
a[1] = 1;
ct += 2;
for(int i = 2; a[i - 1] <= 1000000000; i++){ //条件写为a[i]<=100000000000就崩了，注意是先i++,再判断条件的，所以会死循环
a[i] = a[i - 1] + a[i - 2];
ct++;
}
}
void dfs(int n, int k){
if(n == 1){
flag = 1;
return ;
}
for(int i = k; i >= 3; i--){
if(a[i] > n){
continue;
}
else if(n % a[i] == 0){
dfs(n / a[i], i); //注意：不是k-1而是i
}
if(flag) //剪纸
return ;
}
}
int main(){
std::ios::sync_with_stdio(false);
int t, n;
init();
cin >> t;
while(t--){
cin >> n;
if(n == 0 || n == 1){ //要特判
cout << "Yes" << endl;
}
else{
flag = 0;
dfs(n, ct - 1);
if(flag){
cout << "Yes" << endl;
}
else{
cout << "No" << endl;
}
}
}
return 0;
}