给定两个整数n和k。找到斐波那契数列中K的第n个倍数的位置。
例子：
Input : k = 2, n = 3
Output : 9
3'rd multiple of 2 in Fibonacci Series is 34
which appears at position 9.
Input : k = 4, n = 5
Output : 30
5'th multiple of 4 in Fibonacci Series is 832040
which appears at position 30.
一个有效的解决方案基于以下有趣的属性。
斐波那契数列在模块表示下始终是周期性的。以下是示例。
F (mod 2) = 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0,
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0
Here 0 is repeating at every 3rd index and
the cycle repeats at every 3rd index.
F (mod 3) = 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2
Here 0 is repeating at every 4th index and
the cycle repeats at every 8th index.
F (mod 4) = 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3,
1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0
Here 0 is repeating at every 6th index and
the cycle repeats at every 6th index.
F (mod 5) = 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0,
2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0
Here 0 is repeating at every 5th index and
the cycle repeats at every 20th index.
F (mod 6) = 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4,
3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2
Here 0 is repeating at every 12th index and
the cycle repeats at every 24th index.
F (mod 7) = 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1,
0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6
Here 0 is repeating at every 8th index and
the cycle repeats at every 16th index.
F (mod 8) = 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2,
3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0
Here 0 is repeating at every 6th index and
the cycle repeats at every 12th index.
F (mod 9) = 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7,
6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8
Here 0 is repeating at every 12th index and
the cycle repeats at every 24th index.
F (mod 10) = 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0,
7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0.
Here 0 is repeating at every 15th index and
the cycle repeats at every 60th index.
C++
// C++ program to find position of n'th multiple
// of a mumber k in Fibonacci Series
#include
using namespace std;
const int MAX = 1000;
// Returns position of n'th multple of k in
// Fibonacci Series
int findPosition(int k, int n)
{
// Iterate through all fibonacci numbers
unsigned long long int f1 = 0, f2 = 1, f3;
for (int i = 2; i <= MAX; i++) {
f3 = f1 + f2;
f1 = f2;
f2 = f3;
// Found first multiple of k at position i
if (f2 % k == 0)
// n'th multiple would be at position n*i
// using Periodic property of Fibonacci
// numbers under modulo.
return n * i;
}
}
// Driver Code
int main()
{
int n = 5, k = 4;
cout << "Position of n'th multiple of k"
<< " in Fibonacci Series is "
<< findPosition(k, n) << endl;
return 0;
} 输出：
Position of n'th multiple of k in Fibonacci Series is 30
有关更多详细信息，请参阅斐波那契数列第n个整数倍的完整文章！
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