给我们一个数字N。任务是找到具有N个数字和零的奇数的数字的计数。
注意:该数字可以以0开头。
例子:
Input : N = 2
Output : Count = 18
Input : N = 3
Output : Count = 244
假设一个N位数字仅包含一个零。因此,只能以1种方式填充数字中的数字为零,其余位置可以用9种不同的方式填充从1到9的数字。因此,用N位数字对所有此类数字进行计数,并且只有1个零= N C 1 *(9 N-1 ) 。
类似地,所有具有N位数字和3个零的此类数字的计数= N C 3 *(9 N-3 ) 。
等等。
因此,所有具有N位数字和奇数个零的数字的计数将为,
NC1*(9N-1) + NC3*(9N-3) + NC5*(9N-5) +…….+ NCK*(9N-K)
Where, K is an odd number less than N.
上面的等式可以写成:
(9N)((NC1 * (1/9)) + (NC3 * (1/9^3)) + (NC5 * (1/9^5)) +…
上式可以表示为两个序列的减法,(9 N )* {((1 + x) N- (1-x) N } / 2,其中x = 1/9
等于
(10N - 8N)/2
下面是上述方法的实现:
C++
// C++ program to count numbers with N digits
// which consists of odd number of 0's
#include
using namespace std;
// Function to count Numbers with N digits
// which consists of odd number of 0's
int countNumbers(int N)
{
return (pow(10, N) - pow(8, N)) / 2;
}
// Driver code
int main()
{
int n = 5;
cout << countNumbers(n) << endl;
return 0;
}
Java
// Java program to count numbers
// with N digits which consists
// of odd number of 0's
import java.io.*;
class GFG {
// Function to count Numbers
// with N digits which consists
// of odd number of 0's
static int countNumbers(int N)
{
return (int)(Math.pow(10, N)
- Math.pow(8, N)) / 2;
}
// Driver code
public static void main(String[] args)
{
int n = 5;
System.out.println(countNumbers(n));
}
}
// This code is contributed by Shashank
Python 3
# Python 3 program to count numbers
# with N digits which consists of
# odd number of 0's
# Function to count Numbers with
# N digits which consists of odd
# number of 0's
def countNumbers(N):
return (pow(10, N) - pow(8, N)) // 2
# Driver code
if __name__ == "__main__":
n = 5
print(countNumbers(n))
# This code is contributed
# by ChitraNayal
C#
// C# program to count numbers
// with N digits which consists
// of odd number of 0's
using System;
class GFG {
// Function to count Numbers
// with N digits which consists
// of odd number of 0's
static int countNumbers(int N)
{
return (int)(Math.Pow(10, N)
- Math.Pow(8, N)) / 2;
}
// Driver code
public static void Main()
{
int n = 5;
Console.WriteLine(countNumbers(n));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
PHP
Javascript
输出
33616
注意:答案可能非常大,因此对于大于9的N,请使用模幂。
时间复杂度: O(log n)
辅助空间: O(1)
想要从精选的最佳视频中学习并解决问题,请查看有关从基础到高级C++的C++基础课程以及有关语言和STL的C++ STL课程。要完成从学习语言到DS Algo等的更多准备工作,请参阅“完整面试准备课程” 。