给定整数N ,任务是找到最小的N位数字,即完美的四次幂。
例子:
Input: N = 2
Output: 16
Only valid numbers are 24 = 16
and 34 = 81 but 16 is the minimum.
Input: N = 3
Output: 256
44 = 256
方法:可以观察到,对于N = 1,2,3,… ,该序列将继续像1,16,256,1296,10000,104976,1048576,…的第N个项为pow( ceil((pow(pow(10,(n – 1)),1/4))),4) 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the smallest n-digit
// number which is a perfect fourth power
int cal(int n)
{
double res = pow(ceil((pow(pow(10,
(n - 1)), 1 / 4) )), 4);
return (int)res;
}
// Driver code
int main()
{
int n = 1;
cout << (cal(n));
}
// This code is contributed by Mohit Kumar
Java
// Java implementation of the approach
class GFG
{
// Function to return the smallest n-digit
// number which is a perfect fourth power
static int cal(int n)
{
double res = Math.pow(Math.ceil((
Math.pow(Math.pow(10,
(n - 1)), 1 / 4) )), 4);
return (int)res;
}
// Driver code
public static void main(String[] args)
{
int n = 1;
System.out.println(cal(n));
}
}
// This code is contributed by CodeMech
Python3
# Python3 implementation of the approach
from math import *
# Function to return the smallest n-digit
# number which is a perfect fourth power
def cal(n):
res = pow(ceil( (pow(pow(10, (n - 1)), 1 / 4) ) ), 4)
return int(res)
# Driver code
n = 1
print(cal(n))
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the smallest n-digit
// number which is a perfect fourth power
static int cal(int n)
{
double res = Math.Pow(Math.Ceiling((
Math.Pow(Math.Pow(10,
(n - 1)), 1 / 4) )), 4);
return (int)res;
}
// Driver code
public static void Main()
{
int n = 1;
Console.Write(cal(n));
}
}
// This code is contributed
// by Akanksha_Rai
Javascript
输出:
1