在只允许 2 位数(4 和 7)的序列中找到第 n 个元素
原文:https://www . geesforgeks . org/find-n-th-element-series-2-digits-4-7-允许/
考虑一系列仅由数字 4 和 7 组成的数字。数列中的前几个数字是 4，7，44，47，74，44744，..等等。给定一个数 n，我们需要找到数列中的第 n 个数。 示例:
Input : n = 2
Output : 7
Input : n = 3
Output : 44
Input : n = 5
Output : 74
Input : n = 6
Output : 77
这个想法是基于这样一个事实，即最后一个数字的值在序列中交替。例如，如果第 I 个数字的最后一位是 4，那么第(i-1)个和第(i+1)个数字的最后一位必须是 7。 我们创建一个大小为(n+1)的数组，并将 4 和 7(这两个总是系列的前两个元素)推给它。对于更多的元素，我们检查 1)如果 I 是奇数， arr[I]= arr[I/2] 10+4； 2)如果是偶数， arr[I]= arr[(I/2)-1] 10+7； 最后返回 arr[n]。
C++
// C++ program to find n-th number in a series
// made of digits 4 and 7
#include <bits/stdc++.h>
using namespace std;
// Return n-th number in series made of 4 and 7
int printNthElement(int n)
{
// create an array of size (n+1)
int arr[n+1];
arr[1] = 4;
arr[2] = 7;
for (int i=3; i<=n; i++)
{
// If i is odd
if (i%2 != 0)
arr[i] = arr[i/2]*10 + 4;
else
arr[i] = arr[(i/2)-1]*10 + 7;
}
return arr[n];
}
// Driver code
int main()
{
int n = 6;
cout << printNthElement(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find n-th number in a series
// made of digits 4 and 7
class FindNth
{
// Return n-th number in series made of 4 and 7
static int printNthElement(int n)
{
// create an array of size (n+1)
int arr[] = new int[n+1];
arr[1] = 4;
arr[2] = 7;
for (int i=3; i<=n; i++)
{
// If i is odd
if (i%2 != 0)
arr[i] = arr[i/2]*10 + 4;
else
arr[i] = arr[(i/2)-1]*10 + 7;
}
return arr[n];
}
// main function
public static void main (String[] args)
{
int n = 6;
System.out.println(printNthElement(n));
}
}
Python 3
# Python3 program to find n-th number
# in a series made of digits 4 and 7
# Return n-th number in series made
# of 4 and 7
def printNthElement(n) :
# create an array of size (n + 1)
arr =[0] * (n + 1);
arr[1] = 4
arr[2] = 7
for i in range(3, n + 1) :
# If i is odd
if (i % 2 != 0) :
arr[i] = arr[i // 2] * 10 + 4
else :
arr[i] = arr[(i // 2) - 1] * 10 + 7
return arr[n]
# Driver code
n = 6
print(printNthElement(n))
# This code is contributed by Nikita Tiwari.
C
// C# program to find n-th number in a series
// made of digits 4 and 7
using System;
class GFG
{
// Return n-th number in series made of 4 and 7
static int printNthElement(int n)
{
// create an array of size (n+1)
int []arr = new int[n+1];
arr[1] = 4;
arr[2] = 7;
for (int i = 3; i <= n; i++)
{
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 4;
else
arr[i] = arr[(i / 2) - 1] * 10 + 7;
}
return arr[n];
}
// Driver code
public static void Main ()
{
int n = 6;
Console.Write(printNthElement(n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find n-th
// number in a series
// made of digits 4 and 7
// Return n-th number in
// series made of 4 and 7
function printNthElement($n)
{
// create an array
// of size (n+1)
$arr[1] = 4;
$arr[2] = 7;
for ($i = 3; $i <= $n; $i++)
{
// If i is odd
if ($i % 2 != 0)
$arr[$i] = $arr[$i / 2] *
10 + 4;
else
$arr[$i] = $arr[($i / 2) - 1] *
10 + 7;
}
return $arr[$n];
}
// Driver code
$n = 6;
echo(printNthElement($n));
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to find n-th number in a series
// made of digits 4 and 7
// Return n-th number in series made of 4 and 7
function printNthElement(n) {
// create an array of size (n+1)
var arr = Array(n + 1).fill(0);
arr[1] = 4;
arr[2] = 7;
for (var i = 3; i <= n; i++) {
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 4;
else
arr[i] = arr[(i / 2) - 1] * 10 + 7;
}
return arr[n];
}
// main function
var n = 6;
document.write(printNthElement(n));
// This code is contributed by Princi Singh
</script>
输出:
77
在只允许 2 位数(4 和 7)的数列中寻找第 n 个元素|集合 2 (log(n)方法) 本文由 Roshni Agarwal 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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