最大和最小数字 K 次的乘积形成的数
原文:https://www . geeksforgeeks . org/number-通过添加其最大和最小数字 k 倍的乘积形成/
给定两个整数 N 和 K ，任务是打印其最大值和最小值的乘积 K 次形成的数字。
示例
输入: N = 14，K = 3 输出: 26 解释: M(0)= 14 M(1)= 14+1 * 4 = 18 M(2)= 18+1 * 8 = 26
输入: N = 487，K = 100000000 T3】输出 : 950
接近
- 一个自然的直觉是运行一个循环 K 次，不断更新 n 的值。
- 但是可以观察到，在一些迭代之后，数字的最小值可能为零，并且在此之后，N 永远不会被更新，因为:
M(N+1)= M(N)+0 *(max _ digit) M(N+1)= M(N)
- 因此，我们只需要算出最小数字何时变为 0。
下面是上述方法的实现:
C++
// C++ Code for the above approach
#include <bits/stdc++.h>
using namespace std;
// function that returns the product of
// maximum and minimum digit of N number.
int prod_of_max_min(int n)
{
int largest = 0;
int smallest = 10;
while (n) {
// finds the last digit.
int r = n % 10;
largest = max(r, largest);
smallest = min(r, smallest);
// Moves to next digit
n = n / 10;
}
return largest * smallest;
}
// Function to find the formed number
int formed_no(int N, int K)
{
if (K == 1) {
return N;
}
K--; // M(1) = N
int answer = N;
while (K--) {
int a_current
= prod_of_max_min(answer);
// check if minimum digit is 0
if (a_current == 0)
break;
answer += a_current;
}
return answer;
}
// Driver Code
int main()
{
int N = 487, K = 100000000;
cout << formed_no(N, K) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the formed number
public static int formed_no(int N, int K)
{
if (K == 1)
{
return N;
}
K--; // M(1) = N
int answer = N;
while(K != 0)
{
int a_current = prod_of_max_min(answer);
// Check if minimum digit is 0
if (a_current == 0)
break;
answer += a_current;
}
return answer;
}
// Function that returns the product of
// maximum and minimum digit of N number.
static int prod_of_max_min(int n)
{
int largest = 0;
int smallest = 10;
while(n != 0)
{
// Finds the last digit.
int r = n % 10;
largest = Math.max(r, largest);
smallest = Math.min(r, smallest);
// Moves to next digit
n = n / 10;
}
return largest * smallest;
}
// Driver code
public static void main(String[] args)
{
int N = 487, K = 100000000;
System.out.println(formed_no(N, K));
}
}
// This code is contributed by coder001
Python 3
# Python3 program for the above approach
# Function to find the formed number
def formed_no(N, K):
if (K == 1):
return N
K -= 1 # M(1) = N
answer = N
while (K != 0):
a_current = prod_of_max_min(answer)
# Check if minimum digit is 0
if (a_current == 0):
break
answer += a_current
K -= 1
return answer
# Function that returns the product of
# maximum and minimum digit of N number.
def prod_of_max_min(n):
largest = 0
smallest = 10
while (n != 0):
# Find the last digit.
r = n % 10
largest = max(r, largest)
smallest = min(r, smallest)
# Moves to next digit
n = n // 10
return largest * smallest
# Driver Code
if __name__ == "__main__":
N = 487
K = 100000000
print(formed_no(N, K))
# This code is contributed by chitranayal
C
// C# program for the above approach
using System;
class GFG {
// Function to find the formed number
public static int formed_no(int N, int K)
{
if (K == 1)
{
return N;
}
K--; // M(1) = N
int answer = N;
while(K != 0)
{
int a_current = prod_of_max_min(answer);
// Check if minimum digit is 0
if (a_current == 0)
break;
answer += a_current;
}
return answer;
}
// Function that returns the product of
// maximum and minimum digit of N number.
static int prod_of_max_min(int n)
{
int largest = 0;
int smallest = 10;
while(n != 0)
{
// Finds the last digit.
int r = n % 10;
largest = Math.Max(r, largest);
smallest = Math.Min(r, smallest);
// Moves to next digit
n = n / 10;
}
return largest * smallest;
}
// Driver code
public static void Main(String[] args)
{
int N = 487, K = 100000000;
Console.WriteLine(formed_no(N, K));
}
}
// This code is contributed by Rohit_ranjan
java 描述语言
<script>
// JavaScript Code for the above approach
// Function that returns the product of
// maximum and minimum digit of N number.
function prod_of_max_min(n)
{
var largest = 0;
var smallest = 10;
while (n)
{
// Finds the last digit.
var r = n % 10;
largest = Math.max(r, largest);
smallest = Math.min(r, smallest);
// Moves to next digit
n = parseInt(n / 10);
}
return largest * smallest;
}
// Function to find the formed number
function formed_no(N, K)
{
if (K == 1)
{
return N;
}
K--; // M(1) = N
var answer = N;
while (K--)
{
var a_current = prod_of_max_min(answer);
// Check if minimum digit is 0
if (a_current == 0) break;
answer += a_current;
}
return answer;
}
// Driver Code
var N = 487,
K = 100000000;
document.write(formed_no(N, K) + "<br>");
// This code is contributed by rdtank
</script>
Output:
950
时间复杂度: O(K) 辅助空间: O(1)
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