Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 795 Accepted Submission(s): 213
Problem Description
Following is the recursive definition of Fibonacci sequence:
F
Fi=
0, i = 0
1, i = 1
Fi−1+Fi−2, i > 1
Now we need to check whether a number can be expressed as the product of numbers in the Fibonacci sequence.(一开始没理解题意,其实我觉得这题意也不大好=。 =,题解说是Fib乘积,但和为什么不行呢,路过的可以帮忙解答一下)Input
There is a number T shows there are T test cases below. ( T < 100000 )
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
Note:
1、Fibonacci序列在1~1000000000范围内只有43个数。
2、判断一个数是几个数乘积的递归算法(掌握);
代码(看过题解):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 using namespace std; 6 const int maxn = 100; 7 int f[maxn], tmp[maxn], j, k; 8 void Fibonacci() 9 { 10 f[0] = 0; f[1] = 1; 11 k = 2; 12 while(f[k-1] + f[k-2] <= 1000000000) 13 { 14 f[k] = f[k-1] + f[k-2]; 15 k++; 16 } 17 } 18 bool Judge(int x, int step)//注意这里没有step的话会tle。。。 19 { 20 if(x == 1) return true; 21 for(int i = step; i < j; i++) 22 { 23 if(x%tmp[i] == 0) 24 { 25 if(Judge(x/tmp[i], i)) 26 return true; 27 } 28 } 29 return false; 30 } 31 int main() 32 { 33 Fibonacci(); 34 int T; 35 scanf("%d", &T); 36 for(int i = 0; i < T; i++) 37 { 38 int n; 39 scanf("%d", &n); 40 if(!n) printf("Yes "); 41 else 42 { 43 j = 0; 44 for(int i = 3; i < k; i++) 45 { 46 if(n%f[i] == 0) 47 tmp[j++] = f[i]; 48 } 49 if(Judge(n, 0)) printf("Yes "); 50 else printf("No "); 51 } 52 } 53 return 0; 54 }