算法题（在只允许两位数字(4和7)的序列中查找第n个元素）

本文概述

C ++
Java
Python3
C#
的PHP

【算法题（在只允许两位数字(4和7)的序列中查找第n个元素）】考虑一个仅由数字4和7组成的数字系列。该系列中的前几个数字是4、7、44、47、74、44744等。给定一个数字n, 我们需要在其中找到第n个数字。该序列。
例子：

Input : n = 2 Output : 7Input : n = 3 Output : 44Input: n = 5 Output : 74Input: n = 6 Output : 77

这个想法是基于这样的事实, 即最后一位的值是交替排列的。例如, 如果第i个数字的最后一位为4, 则第(i-1)个和第(i + 1)个数字的最后一位必须为7。
我们创建一个大小为(n + 1)的数组, 然后将其推入4和7(这两个始终是序列的前两个元素)。有关更多元素, 请检查
1)如果是奇数, 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

// Java program to find n-th number in a series // made of digits 4 and 7class 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)); } }

Python3

# 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 ] = 7for 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 + 7return arr[n]# Driver code n = 6 print (printNthElement(n))# This code is contributed by Nikita Tiwari.

// 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.

的PHP

< ?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. ?>

输出如下：