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📜  通过在每一步中减去N的最高有效位来将N减小为零的步骤

📅  最后修改于: 2021-06-26 23:13:05             🧑  作者: Mango

给定一个数字N 。通过在每个步骤中用最高有效数字(最左位数)减去该数字,将其减少为零。任务是计算减少到零所需的步骤数。

例子

Input: 14
Output: 6
Steps:
14 - 1 = 13
13 - 1 = 12
12 - 1 = 11
11 - 1 = 10
10 - 1 = 9
9 - 9 = 0

Input: 20  
Output: 12
Numbers after series of steps:
20, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 0

幼稚的方法:幼稚的方法是逐步减少位数并找到步数,但是如果提供大量数字,则时间复杂度将很大。

高效方法:高效方法的主要思想是减少幼稚方法的步骤数。我们可以跳过前导数字在连续数字中相同的步骤,并对其进行计数。跳过那些具有相同前导数字的数字的算法如下:

  • 让数字为最后一个,将最后一个数字计数,然后将其减少1,因为具有相同前导数字和相同数字计数的最小数字将具有该数字零。
  • 通过last / count查找姓氏数的第一位。
  • 因此,具有相同前导数字的相同位数的最小数目将是[第一位*(count-1)]
  • 跳过的步数可以通过[(最后一个最小数)/第一位数]来实现
  • 因此,下一个数字倒数第二个将是倒数–(第一个*跳过)

下面是上述方法的实现:

C++
// C++ program to find the count of Steps to
// reduce N to zero by subtracting its most
// significant digit at every step
  
#include 
using namespace std;
  
// Function to count the number
// of digits in a number m
int countdig(int m)
{
    if (m == 0)
        return 0;
    else
        return 1 + countdig(m / 10);
}
  
// Function to count the number of
// steps to reach 0
int countSteps(int x)
{
    // count the total number of stesp
    int c = 0;
    int last = x;
  
    // iterate till we reach 0
    while (last) {
  
        // count the digits in last
        int digits = countdig(last);
  
        // decrease it by 1
        digits -= 1;
  
        // find the number on whose division,
        // we get the first digit
        int divisor = pow(10, digits);
  
        // first digit in last
        int first = last / divisor;
  
        // find the first number less than
        // last where the first digit changes
        int lastnumber = first * divisor;
  
        // find the number of numbers
        // with same first digit that are jumped
        int skipped = (last - lastnumber) / first;
  
        skipped += 1;
  
        // count the steps
        c += skipped;
  
        // the next number with a different
        // first digit
        last = last - (first * skipped);
    }
  
    return c;
}
  
// Driver code
int main()
{
    int n = 14;
  
    cout << countSteps(n);
  
    return 0;
}


Java
// Java program to find the count of Steps to
// reduce N to zero by subtracting its most
// significant digit at every step
  
  
class GFG{
// Function to count the number
// of digits in a number m
static int countdig(int m)
{
    if (m == 0)
        return 0;
    else
        return 1 + countdig(m / 10);
}
  
// Function to count the number of
// steps to reach 0
static int countSteps(int x)
{
    // count the total number of stesp
    int c = 0;
    int last = x;
  
    // iterate till we reach 0
    while (last>0) {
  
        // count the digits in last
        int digits = countdig(last);
  
        // decrease it by 1
        digits -= 1;
  
        // find the number on whose division,
        // we get the first digit
        int divisor = (int)Math.pow(10, digits);
  
        // first digit in last
        int first = last / divisor;
  
        // find the first number less than
        // last where the first digit changes
        int lastnumber = first * divisor;
  
        // find the number of numbers
        // with same first digit that are jumped
        int skipped = (last - lastnumber) / first;
  
        skipped += 1;
  
        // count the steps
        c += skipped;
  
        // the next number with a different
        // first digit
        last = last - (first * skipped);
    }
  
    return c;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 14;
  
    System.out.println(countSteps(n));
}
}
// This code is contributed by mits


Python 3
# Python 3 program to find the
# count of Steps to reduce N to
# zero by subtracting its most
# significant digit at every step
  
# Function to count the number
# of digits in a number m
def countdig(m) :
  
    if (m == 0) :
        return 0
    else :
        return 1 + countdig(m // 10)
  
# Function to count the number 
# of steps to reach 0
def countSteps(x) :
      
    # count the total number 
    # of stesp
    c = 0
    last = x
  
    # iterate till we reach 0
    while (last) :
  
        # count the digits in last
        digits = countdig(last)
  
        # decrease it by 1
        digits -= 1
  
        # find the number on whose 
        # division, we get the first digit
        divisor = pow(10, digits)
  
        # first digit in last
        first = last // divisor
  
        # find the first number less 
        # than last where the first 
        # digit changes
        lastnumber = first * divisor
  
        # find the number of numbers
        # with same first digit that 
        # are jumped
        skipped = (last - lastnumber) // first
  
        skipped += 1
  
        # count the steps
        c += skipped
  
        # the next number with a different
        # first digit
        last = last - (first * skipped)
  
    return c
  
# Driver code
n = 14
print(countSteps(n))
  
# This code is contributed by ANKITRAI1


C#
// C# program to find the count of Steps to
// reduce N to zero by subtracting its most
// significant digit at every step
using System;
  
class GFG{
// Function to count the number
// of digits in a number m
static int countdig(int m)
{
    if (m == 0)
        return 0;
    else
        return 1 + countdig(m / 10);
}
  
// Function to count the number of
// steps to reach 0
static int countSteps(int x)
{
    // count the total number of stesp
    int c = 0;
    int last = x;
  
    // iterate till we reach 0
    while (last>0) {
  
        // count the digits in last
        int digits = countdig(last);
  
        // decrease it by 1
        digits -= 1;
  
        // find the number on whose division,
        // we get the first digit
        int divisor = (int)Math.Pow(10, digits);
  
        // first digit in last
        int first = last / divisor;
  
        // find the first number less than
        // last where the first digit changes
        int lastnumber = first * divisor;
  
        // find the number of numbers
        // with same first digit that are jumped
        int skipped = (last - lastnumber) / first;
  
        skipped += 1;
  
        // count the steps
        c += skipped;
  
        // the next number with a different
        // first digit
        last = last - (first * skipped);
    }
  
    return c;
}
  
// Driver code
static void Main()
{
    int n = 14;
  
    Console.WriteLine(countSteps(n));
}
}
// This code is contributed by mits


PHP


输出:
6

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