给定一个数字M，该数字表示要用于打印前N个自然数的最大字符数(忽略空格)。找出N的最大值。
例子：
Input : M = 5
Output : 5
We can type 1 2 3 4 5 using 5 key
presses.
Input : M = 15
Output : 12
We can type 1 2 3 4 5 6 7 8 9 10
11 12 using 15 key presses.
观察到，对于小于11的M，我们可以打印1到9。因此，N将是9。现在，对于10到99之间的数字(总共90个数字)，我们需要两个字符。对于100到999之间的数字(总共900个数字)，我们需要三个字符。因此，请继续计算并从M中减去字符数。此外，当M小于键入的字符数时。查找可以使用允许的其余按键进行打印的偏移号。
以下是此方法的实现：
C/C++
// Maximum natural number that can be printed
// with M characters.
#include
int printMaxN(int m)
{
// At starting point, from 1 to 9, we
// will have only one digit
int total_numbers_within_range = 9;
int number_of_digits = 1;
// While the number of characters is
// greater than the total number of
// natural number in given range e.g.
// if m = 12, then at first step, (m >
// (9)*(1)) evaluates to true
while (m > total_numbers_within_range * number_of_digits) {
// Now here we have exhausted some
// of the digits in making up some natural
// numbers, we reduce the count of m
m = m - (total_numbers_within_range * number_of_digits);
// Increment the number of digits
number_of_digits++;
// Increase the range of the digits
total_numbers_within_range *= 10;
}
// Gives the starting point of any range
int ans = (total_numbers_within_range) / 9 - 1;
// Add the add the remaining digits/(number of
// digits required for current series)
ans += (m / number_of_digits);
return ans;
}
// Driver code
int main()
{
int m = 15;
printf("%dn", printMaxN(m));
return 0;
} Java
// Maximum natural number that
// can be printed with M characters.
import java.util.*;
class GFG {
static int printMaxN(int m)
{
// At starting point, from 1 to 9, we
// will have only one digit
int total_numbers_within_range = 9;
int number_of_digits = 1;
// While the number of characters is
// greater than the total number of
// natural number in given range e.g.
// if m = 12, then at first step, (m >
// (9)*(1)) evaluates to true
while (m > total_numbers_within_range * number_of_digits) {
// Now here we have exhausted some
// of the digits in making up some natural
// numbers, we reduce the count of m
m = m - (total_numbers_within_range * number_of_digits);
// Increment the number of digits
number_of_digits++;
// Increase the range of the digits
total_numbers_within_range *= 10;
}
// Gives the starting point of any range
int ans = (total_numbers_within_range) / 9 - 1;
// Add the add the remaining digits/(number of
// digits required for current series)
ans += (m / number_of_digits);
return ans;
}
// Driver code
public static void main(String[] args)
{
int m = 15;
System.out.print(printMaxN(m));
}
}
// This code is contributed by Anant Agarwal.Python3
# Maximum natural number
# that can be printed
# with M characters.
def printMaxN(m):
# At starting point, from 1 to 9, we
# will have only one digit
total_numbers_within_range = 9
number_of_digits = 1
''' While the number of characters is
greater than the total number of
natural number in given range e.g.'''
# if m = 12, then at first step, (m >
# (9)*(1)) evaluates to true
while (m > total_numbers_within_range * number_of_digits):
# Now here we have exhausted some
# of the digits in making up some natural
# 3 numbers, we reduce the count of m
m = m - (total_numbers_within_range * number_of_digits)
# Increment the number of digits
number_of_digits = number_of_digits + 1
# Increase the range of the digits
total_numbers_within_range = total_numbers_within_range * 10
# Gives the starting point of any range
ans = (total_numbers_within_range) // 9 - 1
# Add the add the remaining digits/(number of
# digits required for current series)
ans = ans + (m // number_of_digits)
return ans
# Driver code
m = 15
print(printMaxN(m))
# This code is contributed
# by Anant Agarwal.C#
// Maximum natural number that
// can be printed with M characters.
using System;
class GFG {
static int printMaxN(int m)
{
// At starting point, from 1 to 9, we
// will have only one digit
int total_numbers_within_range = 9;
int number_of_digits = 1;
// While the number of characters is
// greater than the total number of
// natural number in given range e.g.
// if m = 12, then at first step, (m >
// (9)*(1)) evaluates to true
while (m > total_numbers_within_range * number_of_digits) {
// Now here we have exhausted some
// of the digits in making up some natural
// numbers, we reduce the count of m
m = m - (total_numbers_within_range * number_of_digits);
// Increment the number of digits
number_of_digits++;
// Increase the range of the digits
total_numbers_within_range *= 10;
}
// Gives the starting point of any range
int ans = (total_numbers_within_range) / 9 - 1;
// Add the add the remaining digits/(number of
// digits required for current series)
ans += (m / number_of_digits);
return ans;
}
// Driver code
public static void Main()
{
int m = 15;
Console.Write(printMaxN(m));
}
}
// This code is contributed by vt_m.PHP
// (9)*(1)) evaluates to true
while ($m > $total_numbers_within_range *
$number_of_digits)
{
// Now here we have
// exhausted some of
// the digits in making
// up some natural numbers,
// we reduce the count of m
$m = $m - ($total_numbers_within_range *
$number_of_digits);
// Increment the number
// of digits
$number_of_digits++;
// Increase the range
// of the digits
$total_numbers_within_range *= 10;
}
// Gives the starting
// point of any range
$ans = ($total_numbers_within_range)
/ 9 - 1;
// Add the add the remaining
// digits/(number of digits
// required for current series)
$ans += ($m / $number_of_digits);
return $ans;
}
// Driver Code
$m = 15;
echo printMaxN($m);
// This code is contributed by ajit
?>输出：
12
参考 ：
https://practice.geeksforgeeks.org/problems/faulty-keyboard/0