给定一系列仅由数字3和5组成的数字。该系列中的前几个数字是:
3, 5, 33, 35, 53, 55, …..
给定数字N。任务是找到给定序列中的第n个数字。
例子:
Input : N = 2
Output : 5
Input : N = 5
Output : 53
该想法基于以下事实:序列中最后一位的值交替出现。例如,如果第i个数字的最后一位为3,则第(i-1)个和第(i + 1)个数字的最后一位必须为5。
创建一个大小为(n + 1)的数组,然后将其推入3和5(这两个始终是序列的前两个元素)。有关更多元素的检查,
1) If i is odd,
arr[i] = arr[i/2]*10 + 3;
2) If it is even,
arr[i] = arr[(i/2)-1]*10 + 5;
At last return arr[n].
下面是上述想法的实现:
C++
// C++ program to find n-th number in a series
// made of digits 3 and 5
#include
using namespace std;
// Function to find n-th number in series
// made of 3 and 5
int printNthElement(int n)
{
// create an array of size (n+1)
int arr[n + 1];
arr[1] = 3;
arr[2] = 5;
for (int i = 3; i <= n; i++) {
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
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 3 and 5
class FindNth {
// Function to find n-th number in series
// made of 3 and 5
static int printNthElement(int n)
{
// create an array of size (n+1)
int arr[] = new int[n + 1];
arr[1] = 3;
arr[2] = 5;
for (int i = 3; i <= n; i++) {
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
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 3 and 5
# Return n-th number in series made
# of 3 and 5
def printNthElement(n) :
# create an array of size (n + 1)
arr =[0] * (n + 1);
arr[1] = 3
arr[2] = 5
for i in range(3, n + 1) :
# If i is odd
if (i % 2 != 0) :
arr[i] = arr[i // 2] * 10 + 3
else :
arr[i] = arr[(i // 2) - 1] * 10 + 5
return arr[n]
# Driver code
n = 6
print(printNthElement(n))
C#
// C# program to find n-th number
// in a series made of digits 3 and 5
using System;
class GFG
{
// Function to find n-th number
// in series made of 3 and 5
static int printNthElement(int n)
{
// create an array of size (n+1)
int [] arr = new int[n + 1];
arr[1] = 3;
arr[2] = 5;
for (int i = 3; i <= n; i++)
{
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
return arr[n];
}
// Driver Code
static void Main()
{
int n = 6;
Console.WriteLine(printNthElement(n));
}
}
// This code is contributed by ANKITRAI1
PHP
Javascript
输出:
55
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