执行 p 次运算后检查数组的最后一个元素是偶数还是奇数
原文:https://www . geesforgeks . org/check-last-element-array-偶数-奇数-执行-operation-p-times/
给定的是 n 个非负整数的数组。操作是在数组中插入一个严格大于数组当前总和的数字。执行 p 次运算后，找出数组的最后一个元素是偶数还是奇数。 示例:
Input : arr[] = {2, 3}
P = 3
Output : EVEN
For p = 1, Array sum = 2 + 3 = 5\.
So, we insert 6.
For p = 2, Array sum = 5 + 6 = 11\.
So, we insert 12.
For p = 3, Array sum = 11 + 12 = 23\.
So, we insert 24 (which is even).
Input : arr[] = {5, 7, 10}
p = 1
Output : ODD
For p = 1, Array sum = 5 + 7 + 10 = 22.
So, we insert 23 (which is odd).
天真方法:首先求给定数组的和。这可以在单个循环中完成。现在制作另一个大小为 P + N 的数组。这个数组将表示要插入的元素，最后一个元素将是我们需要的答案。在任何一步，如果数组元素之和的奇偶性为“偶数”，则插入元素的奇偶性为“奇数”。 高效方法:假设数组的和为偶数，下一个插入的元素为奇数。现在数组的和将是奇数，所以下一个插入的元素将是偶数，现在数组的和变成奇数，所以我们插入一个偶数，以此类推。我们可以推广，如果数组的和为偶数，那么对于 P = 1，最后插入的数为奇数，否则为偶数。 现在考虑数组的和为奇数的情况。下一个插入的元素将是偶数，现在数组的和将变成奇数，所以下一个插入的元素将是偶数，现在数组的和将是奇数，再加上另一个偶数，以此类推。在这种情况下，我们可以归纳出最后插入的数字总是偶数。 以下是上述方法的实现:
C++
// CPP program to check whether the last
// element of the array is even or odd
// after performing the operation p times.
#include <bits/stdc++.h>
using namespace std;
string check_last(int arr[], int n, int p)
{
int sum = 0;
// sum of the array.
for (int i = 0; i < n; i++)
sum = sum + arr[i];
if (p == 1) {
// if sum is even
if (sum % 2 == 0)
return "ODD";
else
return "EVEN";
}
return "EVEN";
}
// driver code
int main()
{
int arr[] = { 5, 7, 10 }, p = 1;
int n = sizeof(arr) / sizeof(arr[0]);
cout << check_last(arr, n, p) << endl;
return 0;
}
Python 3
# Python3 code to check whether the last
# element of the array is even or odd
# after performing the operation p times.
def check_last (arr, n, p):
_sum = 0
# sum of the array.
for i in range(n):
_sum = _sum + arr[i]
if p == 1:
# if sum is even
if _sum % 2 == 0:
return "ODD"
else:
return "EVEN"
return "EVEN"
# driver code
arr = [5, 7, 10]
p = 1
n = len(arr)
print(check_last (arr, n, p))
# This code is contributed by "Abhishek Sharma 44"
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check whether the last
// element of the array is even or odd
// after performing the operation p times.
import java.util.*;
class Even_odd{
public static String check_last(int arr[], int n, int p)
{
int sum = 0;
// sum of the array.
for (int i = 0; i < n; i++)
sum = sum + arr[i];
if (p == 1) {
// if sum is even
if (sum % 2 == 0)
return "ODD";
else
return "EVEN";
}
return "EVEN";
}
public static void main(String[] args)
{
int arr[] = { 5, 7, 10 }, p = 1;
int n = 3;
System.out.print(check_last(arr, n, p));
}
}
//This code is contributed by rishabh_jain
C
// C# program to check whether the last
// element of the array is even or odd
// after performing the operation p times.
using System;
class GFG {
public static string check_last(int []arr, int n, int p)
{
int sum = 0;
// sum of the array.
for (int i = 0; i < n; i++)
sum = sum + arr[i];
if (p == 1) {
// if sum is even
if (sum % 2 == 0)
return "ODD";
else
return "EVEN";
}
return "EVEN";
}
// Driver code
public static void Main()
{
int []arr = { 5, 7, 10 };
int p = 1;
int n = arr.Length;
Console.WriteLine(check_last(arr, n, p));
}
}
//This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check whether the last
// element of the array is even or odd
// after performing the operation p times.
function check_last($arr, $n, $p)
{
$sum = 0;
// sum of the array.
for ( $i = 0; $i < $n; $i++)
$sum = $sum + $arr[$i];
if ($p == 1) {
// if sum is even
if ($sum % 2 == 0)
return "ODD";
else
return "EVEN";
}
return "EVEN";
}
// Driver Code
$arr = array(5, 7, 10);
$p = 1;
$n = count($arr);
echo check_last($arr, $n, $p);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript program to check whether the last
// element of the array is even or odd
// after performing the operation p times.
function check_last(arr, n, p)
{
let sum = 0;
// sum of the array.
for (let i = 0; i < n; i++)
sum = sum + arr[i];
if (p == 1) {
// if sum is even
if (sum % 2 == 0)
return "ODD";
else
return "EVEN";
}
return "EVEN";
}
let arr = [ 5, 7, 10 ], p = 1;
let n = arr.length;
document.write(check_last(arr, n, p));
// This code is contributed by divyesh072019.
</script>
输出:
ODD
时间复杂度: O(n)
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