第 n 项，其中 K+1 项是第 K 项与第 K 项的最大和最小位数之差的乘积
原文:https://www . geeksforgeeks . org/n-term-where-k1th-term-是 kth-term-带有最大和最小数字差值的产品-kth-term/
给定两个整数 N 和 D ，任务是求 F(N)的值，其中 F(1)的值为 D，其中 F(K)的给定为:
例:
输入: N = 3，D = 487 输出: 15584 解释: As F(1) = 487， F(2)= 487 (maxDigit(487)–minDigit(487))= 487 * 4 = 1948 F(3)= 1948 (maxDigit(1948)–minDigit(1948
方法:思路是借助循环迭代计算 F(2)到 F(N)的值。此外，每个数字中的最大和最小位数可以通过将数字除以 10 来计算，同时取模得到位数。 以下是上述方法的实施:
C++
// C++ implementation to find the value
// of the given function for the value
#include <bits/stdc++.h>
using namespace std;
// Function to find minimum digit
// in the decimal representation of N
int MIN(int n)
{
int ans = 11;
// Loop to find the minimum
// digit in the number
while (n) {
ans = min(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find maximum digit
// in the decimal representation of N
int MAX(int n)
{
int ans = -1;
// Loop to find the maximum
// digit in the number
while (n) {
ans = max(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find the value
// of the given function
void Find_value(int n, int k)
{
k--;
int x = 0;
int y = 0;
// Loop to compute the values
// of the given number
while (k--) {
x = MIN(n);
y = MAX(n);
if (y - x == 0)
break;
n *= (y - x);
}
cout << n;
}
// Driver Code
int main()
{
int N = 487, D = 5;
// Function Call
Find_value(N, D);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find the value
// of the given function for the value
class GFG{
// Function to find minimum digit in
// the decimal representation of N
static int MIN(int n)
{
int ans = 11;
// Loop to find the minimum
// digit in the number
while (n > 0)
{
ans = Math.min(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find maximum digit
// in the decimal representation of N
static int MAX(int n)
{
int ans = -1;
// Loop to find the maximum
// digit in the number
while (n > 0)
{
ans = Math.max(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find the value
// of the given function
static void Find_value(int n, int k)
{
k--;
int x = 0;
int y = 0;
// Loop to compute the values
// of the given number
while (k-- > 0)
{
x = MIN(n);
y = MAX(n);
if (y - x == 0)
break;
n *= (y - x);
}
System.out.print(n);
}
// Driver Code
public static void main(String[] args)
{
int N = 487, D = 5;
// Function Call
Find_value(N, D);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation to find the value
# of the given function for the value
# Function to find minimum digit
# in the decimal representation of N
def MIN(n):
ans = 11
# Loop to find the minimum
# digit in the number
while n:
ans = min(ans, n % 10)
n //= 10
return ans
# Function to find maximum digit in
# the decimal representation of N
def MAX(n):
ans = -1
# Loop to find the maximum
# digit in the number
while n:
ans = max(ans, n % 10)
n //= 10
return ans
# Function to find the value
# of the given function
def Find_value(n, k):
k -= 1
(x, y) = (0, 0)
# Loop to compute the values
# of the given number
while k:
k -= 1
x = MIN(n)
y = MAX(n)
if ((y - x) == 0):
break
n *= (y - x)
print(n, end = ' ')
# Driver Code
if __name__=='__main__':
(N, D) = (487, 5)
# Function Call
Find_value(N, D)
# This code is contributed by rutvik_56
C
// C# implementation to find the value
// of the given function for the value
using System;
class GFG{
// Function to find minimum digit in
// the decimal representation of N
static int MIN(int n)
{
int ans = 11;
// Loop to find the minimum
// digit in the number
while (n > 0)
{
ans = Math.Min(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find maximum digit
// in the decimal representation of N
static int MAX(int n)
{
int ans = -1;
// Loop to find the maximum
// digit in the number
while (n > 0)
{
ans = Math.Max(ans, n % 10);
n /= 10;
}
return ans;
}
// Function to find the value
// of the given function
static void Find_value(int n, int k)
{
k--;
int x = 0;
int y = 0;
// Loop to compute the values
// of the given number
while (k-- > 0)
{
x = MIN(n);
y = MAX(n);
if (y - x == 0)
break;
n *= (y - x);
}
Console.Write(n);
}
// Driver Code
public static void Main(String[] args)
{
int N = 487, D = 5;
// Function Call
Find_value(N, D);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// javascript implementation to find the value
// of the given function for the value
// Function to find minimum digit
// in the decimal representation of N
function MIN( n)
{
let ans = 11;
// Loop to find the minimum
// digit in the number
while (n) {
ans = parseInt(Math.min(ans, n % 10));
n = parseInt(n/ 10);
}
return ans;
}
// Function to find maximum digit
// in the decimal representation of N
function MAX( n)
{
let ans = -1;
// Loop to find the maximum
// digit in the number
while (n) {
ans = parseInt(Math.max(ans, n % 10));
n = parseInt(n/ 10);
}
return ans;
}
// Function to find the value
// of the given function
function Find_value( n, k)
{
k--;
let x = 0;
let y = 0;
// Loop to compute the values
// of the given number
while (k--) {
x = parseInt(MIN(n));
y = parseInt(MAX(n));
if (y - x == 0)
break;
n *= (y - x);
}
document.write(n);
}
// Driver Code
let N = 487, D = 5;
// Function Call
Find_value(N, D);
// This code contributed by gauravrajput1
</script>
Output:
981792
时间复杂度: O(D*log(N))
辅助空间: O(1)
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