📜  计算满足给定条件的所有可能的N位数字

📅  最后修改于: 2021-06-26 13:43:13             🧑  作者: Mango

给定一个整数N,任务是要计算所有可能的N-位数字,使得A +反向(A)= 10 N – 1其中A是N位数字和反向(A)是反向A. A的不应该有任何前导0。
例子:

方法:首先我们必须得出结论,如果N为奇数,则没有数字可以满足给定条件,让我们针对N = 3进行证明。

现在找到N偶数时的答案。例如N = 4

下面是上述方法的实现:

C++
// C++ implementation of above approach
#include 
using namespace std;
 
// Function to return the count of required numbers
string getCount(int N)
{
 
    // If N is odd then return 0
    if (N % 2 == 1)
        return 0;
 
    string result = "9";
 
    for (int i = 1; i <= N / 2 - 1; i++)
        result += "0";
 
    return result;
}
 
// Driver Code
int main()
{
 
    int N = 4;
    cout << getCount(N);
 
    return 0;
}


Java
// Java implementation of above approach
class GFG
{
    // Function to return the count of required numbers
    static String getCount(int N)
    {
     
        // If N is odd then return 0
        if (N % 2 == 1)
            return "0";
     
        String result = "9";
        for (int i = 1; i <= N / 2 - 1; i++)
            result += "0";
        return result;
    }
     
    // Driver Code
    public static void main(String []args)
    {
     
        int N = 4;
        System.out.println(getCount(N));
    }
}
 
// This code is contributed by ihritik


Python3
# Python3 implementation of above approach
 
# Function to return the count of required numbers
def getCount(N):
 
    # If N is odd then return 0
    if (N % 2 == 1):
        return "0"
 
    result = "9"
 
    for i in range (1, N // 2 ):
        result = result + "0"
 
    return result
 
# Driver Code
N = 4
print(getCount(N))
 
# This code is contributed by ihritik


C#
// C# implementation of above approach
using System;
 
class GFG
{
    // Function to return the count of required numbers
    static string getCount(int N)
    {
     
        // If N is odd then return 0
        if (N % 2 == 1)
            return "0";
        string result = "9";
        for (int i = 1; i <= N / 2 - 1; i++)
            result += "0";
        return result;
    }
     
    // Driver Code
    public static void Main()
    {
     
        int N = 4;
        Console.WriteLine(getCount(N));
    }
}
 
// This code is contributed by ihritik


PHP


输出:
90

时间复杂度: O(N)

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