两个数除法小数点前的位数
给定两个整数 a 和 b 。任务是在 a / b 中找到小数点前的位数。 举例:
输入: a = 100，b = 4 输出: 2 100 / 4 = 25，25 中的位数= 2。 输入: a = 100000，b = 10 输出: 5
天真的做法:将两个数相除，然后求除法中的位数。取除法的绝对值求位数。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
int main()
{
int a = 100;
int b = 10;
cout << countDigits(a, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the number of digits
// before the decimal in a / b
static int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = Math.abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
public static void main(String args[])
{
int a = 100;
int b = 10;
System.out.print(countDigits(a, b));
}
}
计算机编程语言
# Python 3 implementation of the approach
# Function to return the number of digits
# before the decimal in a / b
def countDigits(a, b):
count = 0
# Absolute value of a / b
p = abs(a // b)
# If result is 0
if (p == 0):
return 1
# Count number of digits in the result
while (p > 0):
count = count + 1
p = p // 10
# Return the required count of digits
return count
# Driver code
a = 100
b = 10
print(countDigits(a, b))
C
// C# implementation of the approach
using System;
class GFG {
// Function to return the number of digits
// before the decimal in a / b
static int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = Math.Abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
public static void Main()
{
int a = 100;
int b = 10;
Console.Write(countDigits(a, b));
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits($a, $b)
{
$count = 0;
// Absolute value of a / b
$p = abs($a / $b);
// If result is 0
if ($p == 0)
return 1;
// Count number of digits in the result
while ($p > 0) {
$count++;
$p = (int)($p / 10);
}
// Return the required count of digits
return $count;
}
// Driver code
$a = 100;
$b = 10;
echo countDigits($a, $b);
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits(a, b)
{
var count = 0;
// Absolute value of a / b
var p = Math.abs(parseInt(a / b));
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = parseInt(p / 10);
}
// Return the required count of digits
return count;
}
// Driver code
var a = 100;
var b = 10;
document.write(countDigits(a, b));
// This code is contributed by rrrtnx.
</script>
Output:
2
有效方法:要计算 a / b 中的位数，我们可以使用公式:
楼层(原木10(a)–原木 10 (b)) + 1
这里两个数字都必须是正整数。为此我们可以取 a 和 b 的绝对值。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
// Return the required count of digits
return floor(log10(abs(a)) - log10(abs(b))) + 1;
}
// Driver code
int main()
{
int a = 100;
int b = 10;
cout << countDigits(a, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the number of digits
// before the decimal in a / b
public static int countDigits(int a, int b)
{
double digits = Math.log10(Math.abs(a))
- Math.log10(Math.abs(b)) + 1;
// Return the required count of digits
return (int)Math.floor(digits);
}
// Driver code
public static void main(String[] args)
{
int a = 100;
int b = 10;
System.out.print(countDigits(a, b));
}
}
计算机编程语言
# Python3 implementation of the approach
import math
# Function to return the number of digits
# before the decimal in a / b
def countDigits(a, b):
# Return the required count of digits
return math.floor(math.log10(abs(a)) -
math.log10(abs(b))) + 1
# Driver code
a = 100
b = 10
print(countDigits(a, b))
C
// C# implementation of the approach
using System;
class GFG {
// Function to return the number of digits
// before the decimal in a / b
public static int countDigits(int a, int b)
{
double digits = Math.Log10(Math.Abs(a))
- Math.Log10(Math.Abs(b)) + 1;
// Return the required count of digits
return (int)Math.Floor(digits);
}
// Driver code
static void Main()
{
int a = 100;
int b = 10;
Console.Write(countDigits(a, b));
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits($a, $b)
{
// Return the required count of digits
return floor(log10(abs($a)) -
log10(abs($b))) + 1;
}
// Driver code
$a = 100;
$b = 10;
echo countDigits($a, $b);
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits(a, b)
{
// Return the required count of digits
return Math.floor((Math.log(Math.abs(a))/Math.log(10)) - (Math.log(Math.abs(b))/Math.log(10))) + 1;
}
// Driver code
var a = 100;
var b = 10;
document.write(countDigits(a, b));
// This code is contributed by rutvik_56.
</script>
Output:
2
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