找出只能用数字 3、4 构成且最大长度为 n 的数字的个数
原文:https://www . geesforgeks . org/find-可使用数字 3-4 形成的数字计数-仅限最大长度为-n/
给定一个数 n，找出这些数的个数，这些数只能用数字 3 和 4 构成，长度最大为 n 例:
Input : N = 2
Output : 6
Explanation : 3, 4, 33, 34, 43, 44
are numbers having length 2 and digits 3 and 4 only.
Input : N = 1
Output : 2
Explanation : 3, 4 are the only such numbers.
接近:长度 1 有 2 个数字。他们分别是 3 岁和 4 岁。长度为 2 的数字有 4 个。他们是 33，34，43 和 44。长度为 3 的数字有 8 个。它们是 333，334，343，344，433，434，443，444。长度每增加 1，数字就会增加 2 倍。 很容易证明:对于前一长度的任何数字，可以附加 3 或 4，所以前一长度的一个数字产生下一长度的两个数字。 那么对于长度 N，这样的长度 N 的数的量正好是 2*N，但是在问题中，我们需要长度不大于 N 的数的个数，我们来总结一下。2 1 = 2，2 1 + 2 2 = 2 + 4 = 6，21+22+23= 2+4+8 = 14，21+22+23+24= 2+4+4 人们可以注意到，之前所有 2 的幂的和等于下一个 2 的幂减去第一个 2 的幂。所以问题的答案是 2N+1–2。 以下是上述方法的实施:
C++
// Cpp program to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
long long numbers(int n)
{
return (long long)(pow(2, n + 1)) - 2;
}
// Driver code
int main()
{
int n = 2;
cout << numbers(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
class GFG
{
// Function to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
static long numbers(int n)
{
return (long)(Math.pow(2, n + 1)) - 2;
}
// Driver code
public static void main(String args[])
{
int n = 2;
System.out.println( numbers(n));
}
}
// This code is contributed by Arnab Kundu
Python 3
# Python3 program to find the count of
# numbers that can be formed using digits
# 3, 4 only and having length at max N.
# Function to find the count of numbers
# that can be formed using digits 3, 4
# only and having length at max N.
def numbers(n):
return pow(2, n + 1) - 2
# Driver code
n = 2
print(numbers(n))
# This code is contributed
# by Shrikant13
C
// C# program to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
using System;
class GFG
{
// Function to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
static long numbers(int n)
{
return (long)(Math.Pow(2, n + 1)) - 2;
}
// Driver code
static void Main()
{
int n = 2;
Console.WriteLine( numbers(n));
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the count of
// numbers that can be formed using
// digits 3, 4 only and having length
// at max N.
// Function to find the count of numbers
// that can be formed using digits 3, 4 only
// and having length at max N.
function numbers($n)
{
return (pow(2, $n + 1)) - 2;
}
// Driver code
$n = 2;
echo numbers($n);
// This code is contributed
// by Akanksha Rai
?>
java 描述语言
<script>
// javascript program to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
// Function to find the count of numbers that
// can be formed using digits 3, 4 only and
// having length at max N.
function numbers(n) {
return (Math.pow(2, n + 1)) - 2;
}
// Driver code
var n = 2;
document.write(numbers(n));
// This code is contributed by Princi Singh
</script>
Output:
6
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