检查偶数和奇数位数字的乘积是否相等
给定一个整数 N ，任务是检查一个数的偶数和奇数位的数字乘积是否相等。如果相等，打印是否则打印否。
示例:
输入: N = 2841 输出:是 奇数位位数乘积= 2 * 4 = 8 偶数位位数乘积= 8 * 1 = 8
输入: N = 4324 输出: No 奇数位数字乘积= 4 * 2 = 8 偶数位数字乘积= 3 * 4 = 12
进场:
- 在偶数位置找到数字的乘积并存储在prod ever中。
- 在奇数位置找到数字的乘积并存储在 prodOdd 中。
- 如果 prodEven = prodOdd 则打印是否则打印否。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if the product
// of even positioned digits is equal to
// the product of odd positioned digits in n
bool productEqual(int n)
{
// If n is a single digit number
if (n < 10)
return false;
int prodOdd = 1, prodEven = 1;
while (n > 0) {
// Take two consecutive digits
// at a time
// last digit
int digit = n % 10;
prodEven *= digit;
n /= 10;
// Second last digit
digit = n % 10;
prodOdd *= digit;
n /= 10;
}
// If the products are equal
if (prodEven == prodOdd)
return true;
// If products are not equal
return false;
}
// Driver code
int main()
{
int n = 4324;
if (productEqual(n))
cout << "Yes";
else
cout << "No";
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 that returns true
// if the product of even positioned
// digits is equal to the product of
// odd positioned digits in n
static boolean productEqual(int n)
{
// If n is a single digit number
if (n < 10)
return false;
int prodOdd = 1, prodEven = 1;
while (n > 0) {
// Take two consecutive digits
// at a time
// First digit
int digit = n % 10;
prodOdd *= digit;
n /= 10;
// If n becomes 0 then
// there's no more digit
if (n == 0)
break;
// Second digit
digit = n % 10;
prodEven *= digit;
n /= 10;
}
// If the products are equal
if (prodEven == prodOdd)
return true;
// If products are not equal
return false;
}
// Driver code
public static void main(String args[])
{
int n = 4324;
if (productEqual(n))
System.out.println("Yes");
else
System.out.println("No");
}
// This code is contributed by Ryuga
}
Python 3
# Python implementation of the approach
# Function that returns true if the product
# of even positioned digits is equal to
# the product of odd positioned digits in n
def productEqual(n):
if n < 10:
return False
prodOdd = 1
prodEven = 1
# Take two consecutive digits
# at a time
# First digit
while n > 0:
digit = n % 10
prodOdd *= digit
n = n//10
# If n becomes 0 then
# there's no more digit
if n == 0:
break
digit = n % 10
prodEven *= digit
n = n//10
# If the products are equal
if prodOdd == prodEven:
return True
# If the products are not equal
return False
# Driver code
n = 4324
if productEqual(n):
print("Yes")
else:
print("No")
# This code is contributed by Shrikant13
C
// C# implementation of the approach
using System;
class GFG {
// Function that returns true
// if the product of even positioned
// digits is equal to the product of
// odd positioned digits in n
static bool productEqual(int n)
{
// If n is a single digit number
if (n < 10)
return false;
int prodOdd = 1, prodEven = 1;
while (n > 0) {
// Take two consecutive digits
// at a time
// First digit
int digit = n % 10;
prodOdd *= digit;
n /= 10;
// If n becomes 0 then
// there's no more digit
if (n == 0)
break;
// Second digit
digit = n % 10;
prodEven *= digit;
n /= 10;
}
// If the products are equal
if (prodEven == prodOdd)
return true;
// If products are not equal
return false;
}
// Driver code
static void Main()
{
int n = 4324;
if (productEqual(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that returns true if the product
// of even positioned digits is equal to
// the product of odd positioned digits in n
function productEqual($n)
{
// If n is a single digit number
if ($n < 10)
return false;
$prodOdd = 1;
$prodEven = 1;
while ($n > 0)
{
// Take two consecutive digits
// at a time
// First digit
$digit = $n % 10;
$prodOdd *= $digit;
$n /= 10;
// If n becomes 0 then
// there's no more digit
if ($n == 0)
break;
// Second digit
$digit = $n % 10;
$prodEven *= $digit;
$n /= 10;
}
// If the products are equal
if ($prodEven == $prodOdd)
return true;
// If products are not equal
return false;
}
// Driver code
$n = 4324;
if (productEqual(!$n))
echo "Yes";
else
echo "No";
// This code is contributed by jit_t
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function that returns true if the product
// of even positioned digits is equal to
// the product of odd positioned digits in n
function productEqual(n)
{
// If n is a single digit number
if (n < 10)
return false;
let prodOdd = 1, prodEven = 1;
while (n > 0) {
// Take two consecutive digits
// at a time
// First digit
let digit = n % 10;
prodOdd *= digit;
n = Math.floor(n / 10);
// If n becomes 0 then
// there's no more digit
if (n == 0)
break;
// Second digit
digit = n % 10;
prodEven *= digit;
n = Math.floor(n / 10);
}
// If the products are equal
if (prodEven == prodOdd)
return true;
// If products are not equal
return false;
}
// Driver code
let n = 4324;
if (productEqual(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by Surbhi Tyagi.
</script>
Output:
No
方法 2:将整数转换为字符串:
- 将整数转换为字符串。遍历字符串，将所有偶数索引的乘积存储在一个变量中，将所有奇数索引的乘积存储在另一个变量中。
- 如果两者相等，则打印是，否则打印否
下面是实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
void getResult(int n)
{
// To store the respective product
int proOdd = 1;
int proEven = 1;
// Converting integer to string
string num = to_string(n);
// Traversing the string
for(int i = 0; i < num.size(); i++)
if (i % 2 == 0)
proOdd = proOdd * (num[i] - '0');
else
proEven = proEven * (num[i] - '0');
if (proOdd == proEven)
cout << "Yes";
else
cout << "No";
}
// Driver code
int main()
{
int n = 4324;
getResult(n);
return 0;
}
// This code is contributed by sudhanshugupta2019a
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG{
static void getResult(int n)
{
// To store the respective product
int proOdd = 1;
int proEven = 1;
// Converting integer to String
String num = String.valueOf(n);
// Traversing the String
for(int i = 0; i < num.length(); i++)
if (i % 2 == 0)
proOdd = proOdd * (num.charAt(i) - '0');
else
proEven = proEven * (num.charAt(i) - '0');
if (proOdd == proEven)
System.out.print("Yes");
else
System.out.print("No");
}
// Driver code
public static void main(String[] args)
{
int n = 4324;
getResult(n);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
def getResult(n):
# To store the respective product
proOdd = 1
proEven = 1
# Converting integer to string
num = str(n)
# Traversing the string
for i in range(len(num)):
if(i % 2 == 0):
proOdd = proOdd*int(num[i])
else:
proEven = proEven*int(num[i])
if(proOdd == proEven):
print("Yes")
else:
print("No")
# Driver code
if __name__ == "__main__":
n = 4324
getResult(n)
# This code is contributed by vikkycirus
C
// C# implementation of the approach
using System;
public class GFG{
static void getResult(int n)
{
// To store the respective product
int proOdd = 1;
int proEven = 1;
// Converting integer to String
String num = String.Join("",n);
// Traversing the String
for(int i = 0; i < num.Length; i++)
if (i % 2 == 0)
proOdd = proOdd * (num[i] - '0');
else
proEven = proEven * (num[i] - '0');
if (proOdd == proEven)
Console.Write("Yes");
else
Console.Write("No");
}
// Driver code
public static void Main(String[] args)
{
int n = 4324;
getResult(n);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
function getResult(n)
{
// To store the respective product
let proOdd = 1;
let proEven = 1;
// Converting integer to String
let num = n.toString();
// Traversing the String
for(let i = 0; i < num.length; i++)
if (i % 2 == 0)
proOdd = proOdd * (num[i].charCodeAt() - '0'.charCodeAt());
else
proEven = proEven * (num[i].charCodeAt() - '0'.charCodeAt());
if (proOdd == proEven)
document.write("Yes");
else
document.write("No");
}
let n = 4324;
getResult(n);
</script>
输出:
No
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