程序以整数形式对数字进行计数(4种不同的方法)

计算用户输入的长整数中的位数。

简单的迭代解决方案
用户输入的整数存储在变量n中。然后，迭代while循环，直到将测试表达式n！= 0评估为0(假)。

1. 第一次迭代后，n的值为345，并且计数增加到1。
2. 在第二次迭代后，n的值为34，并且计数增加到2。
3. 在第三次迭代之后，n的值为3，并且计数增加到3。
4. 在第四次迭代开始时，n的值将为0，并且循环终止。

然后，将测试表达式评估为false并终止循环。

C++

// Iterative C++ program to count
// number of digits in a number
#include 
using namespace std;
 
int countDigit(long long n)
{
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits : " << countDigit(n);
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// Iterative C program to count number of
// digits in a number
#include 
 
int countDigit(long long n)
{
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
class GFG {
 
    static int countDigit(long n)
    {
        int count = 0;
        while (n != 0) {
            n = n / 10;
            ++count;
        }
        return count;
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Iterative Python program to count
# number of digits in a number
 
 
def countDigit(n):
    count = 0
    while n != 0:
        n //= 10
        count += 1
    return count
 
 
# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
 
# This code is contributed by Shreyanshi Arun

C#

// C# Code to count number of
// digits in an integer
using System;
 
class GFG {
 
    static int countDigit(long n)
    {
        int count = 0;
        while (n != 0)
        {
            n = n / 10;
            ++count;
        }
        return count;
    }
 
    /* Driver code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of"
                          + " digits : " + countDigit(n));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// Iterative Javascript program to count
// number of digits in a number
 
function countDigit(n)
{
    let count = 0;
    while (n != 0)
    {
        n = Math.floor(n / 10);
        ++count;
    }
    return count;
}
 
// Driver code
 
    n = 345289467;
    document.write("Number of digits : "+ countDigit(n));
     
// This code is contributed by Mayank Tyagi
 

输出

Number of digits : 9

递归解决方案：

C++

// Recursive C++ program to count number of
// digits in a number
#include 
using namespace std;
 
int countDigit(long long n)
{
    if (n == 0)
        return 0;
    return 1 + countDigit(n / 10);
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits :" << countDigit(n);
    return 0;
}
// This code is contributed by Mukul Singh.

C

// Recursive C program to count number of
// digits in a number
#include 
 
int countDigit(long long n)
{
    if (n == 0)
        return 0;
    return 1 + countDigit(n / 10);
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
import java.util.*;
 
class GFG {
 
    static int countDigit(long n)
    {
        if (n == 0)
            return 0;
        return 1 + countDigit(n / 10);
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# Recursive Python program to count
# number of digits in a number
 
 
def countDigit(n):
    if n == 0:
        return 0
    return 1 + countDigit(n // 10)
 
 
# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
 
# This code is contributed by Shreyanshi Arun

C＃

// C# Code to count number of
// digits in an integer
using System;
 
class GFG {
 
    static int countDigit(long n)
    {
        if (n == 0)
            return 0;
        return 1 + countDigit(n / 10);
    }
 
    /* Driver Code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of "
                          + "digits : "
                          + countDigit(n));
    }
}
 
// This code is contributed by anuj_67.

Java脚本

 
// Recursive Javascript program to count number of
// digits in a number
 
function countDigit(n)
{
    if (n == 0)
        return 0;
    return 1 + countDigit(parseInt(n / 10));
}
 
// Driver code
var n = 345289467;
document.write("Number of digits :" + countDigit(n));
 

输出

Number of digits :9

基于日志的解决方案：
我们可以使用log10(以10为底的对数)来计算正数的位数(未为负数定义对数)。
N的数字计数= log10(N)的上限。

C++

// Log based C++ program to count number of
// digits in a number
#include 
using namespace std;
 
int countDigit(long long n) {
  return floor(log10(n) + 1);
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits : "
         << countDigit(n);
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

// Log based C program to count number of
// digits in a number
#include 
#include 
 
int countDigit(long long n) {
  return floor(log10(n) + 1);
}
 
// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
import java.util.*;
 
class GFG {
 
    static int countDigit(long n)
    {
        return (int)Math.floor(Math.log10(n) + 1);
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Log based Python program to count number of
# digits in a number
 
# function to import ceil and log
import math
 
 
def countDigit(n):
    return math.floor(math.log10(n)+1)
 
 
# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
 
# This code is contributed by Shreyanshi Arun

C＃

// C# Code to count number of
// digits in an integer
using System;
 
class GFG {
 
    static int countDigit(long n)
    {
        return (int)Math.Floor(Math.Log10(n) + 1);
    }
 
    /* Driver code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of digits : "
                          + countDigit(n));
    }
}
 
// This code is contributed by anuj_67..

Java脚本

 
// Log based Javascript program to count number of
// digits in a number
function countDigit(n)
{
    return Math.floor(Math.log10(n) + 1);
}
 
// Driver code
var n = 345289467;
document.write("Number of digits : "  +
               countDigit(n));
                
// This code is contributed by noob2000
 

输出

Number of digits : 9

方法4：
我们可以将数字转换成字符串，然后找到字符串的长度来获得在原号码的位数。

C++

#include 
using namespace std;
 
// To count the no. of digits in a number
void count_digits(int n)
{
    // converting number to string using
    // to_string in C++
    string num = to_string(n);
 
    // calculate the size of string
    cout << num.size() << endl;
}
//Driver Code
int main()
{
    // number
    int n = 345;
    count_digits(n);
    return 0;
}
 
// This code is contributed by Shashank Pathak

Java

import java.util.*;
public class GFG {
 
    // To count the no. of digits in a number
    static void count_digits(int n)
    {
        // converting number to string using
        // to_string in C++
        String num = Integer.toString(n);
 
        // calculate the size of string
 
        System.out.println(+num.length());
    }
    // Driver code
    public static void main(String args[])
    {
        // number
        int n = 345;
        count_digits(n);
    }
}
// Code is contributed by shivanisinghss2110

Python3

# Python3 implementation of the approach
def count_digits(n):
    n = str(n)
    return len(n)
 
 
# Driver code
n = 456533457776
print(count_digits(n))

C＃

// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
    // To count the no. of digits in a number
    static void count_digits(int n)
    {
        // converting number to string using
        // to_string in C#
 
        string num = Convert.ToString(n);
 
        // calculate the size of string
        Console.WriteLine(+num.Length);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        // number
        int n = 345;
        count_digits(n);
    }
}
 
// This code is contributed by shivanisinghss2110

输出

3