检查数字是偶数还是奇数，给出数字和基数
原文:https://www . geeksforgeeks . org/check-如果给定的数字是偶数或奇数，则取其位数和基数的基数/
给定一个大小为 N 的数组arr【】，该数组代表一个数字的位数和一个整数 r ，该整数是给定数字的基数(基数)，即N = arr[N–1] r0+arr[N–2] r1+…+a[0]* rN–1。任务是找出给定的数字是奇数还是偶数。 举例:
输入: arr[] = {1，0}，r = 2 输出:偶数 (10)2=(2)10 T11】输入: arr[] = {1，2，3，4，5，6，7，8，9}，r = 10 输出:奇数
天真法:最简单的方法是将数字乘以相应的基数幂来计算数字 n。但是由于位数可以是 10 5 的数量级，这种方法不适用于大型 n. 高效方法:有两种可能。
- 如果 r 为偶数，则最终答案取决于最后一位数字，即arr[n–1]。
- 如果 r 是奇数那么我们要数奇数位数。如果奇数位数是偶数，那么和就是偶数。否则总和就是奇数。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function that returns true if the number
// represented by arr[] is even in base r
bool isEven(int arr[], int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0) {
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else {
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
int main()
{
int arr[] = { 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
int r = 2;
if (isEven(arr, n, r))
cout << "Even";
else
cout << "Odd";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function that returns true if the number
// represented by arr[] is even in base r
static boolean isEven(int arr[], int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else {
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 0 };
int n = arr.length;
int r = 2;
if (isEven(arr, n, r))
System.out.println ("Even");
else
System.out.println("Odd");
}
}
// This code is contributed by jit_t.
Python 3
# Python 3 implementation of the approach
# Function that returns true if the number
# represented by arr[] is even in base r
def isEven(arr, n, r):
# If the base is even, then
# the last digit is checked
if (r % 2 == 0):
if (arr[n - 1] % 2 == 0):
return True
# If base is odd, then the
# number of odd digits are checked
else:
# To store the count of odd digits
oddCount = 0
for i in range(n):
if (arr[i] % 2 != 0):
oddCount += 1
if (oddCount % 2 == 0):
return True
# Number is odd
return False
# Driver code
if __name__ == '__main__':
arr = [1, 0]
n = len(arr)
r = 2
if (isEven(arr, n, r)):
print("Even")
else:
print("Odd")
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if the number
// represented by arr[] is even in base r
static bool isEven(int []arr, int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i)
{
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 0 };
int n = arr.Length;
int r = 2;
if (isEven(arr, n, r))
Console.WriteLine ("Even");
else
Console.WriteLine("Odd");
}
}
// This code is contributed by anuj_67...
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 number
// represented by arr[] is even in base r
function isEven($arr, $n, $r)
{
// If the base is even, then
// the last digit is checked
if ($r % 2 == 0)
{
if ($arr[$n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
$oddCount = 0;
for ($i = 0; $i < $n; ++$i)
{
if ($arr[$i] % 2 != 0)
$oddCount++;
}
if ($oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
$arr = array( 1, 0 );
$n = Count($arr);
$r = 2;
if (isEven($arr, $n, $r))
echo "Even";
else
echo "Odd";
// This code is contributed by andrew1234
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns true if the number
// represented by arr[] is even in base r
function isEven(arr, n, r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
let oddCount = 0;
for (let i = 0; i < n; ++i)
{
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
let arr = [ 1, 0 ];
let n = arr.length;
let r = 2;
if (isEven(arr, n, r))
document.write("Even");
else
document.write("Odd");
</script>
Output:
Even
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