比较数组的第 N-1 个元素和第 N 个元素的和
原文:https://www . geeksforgeeks . org/compare-first-n-1 元素之和与第 n 个数组元素之比/
给定一个大小为 N 的数组 arr[] ，任务是检查数组的第一个N–1元素的和是否等于最后一个元素。 举例:
输入: arr[] = {1，2，3，4，10} 输出:是 输入: arr[] = {1，2，3，4，12} 输出:否
方法:求数组中第一个N–1元素的和，即arr[0]+arr[1]+…+arr[N–2]并与arr[N–1]进行比较。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
bool isSumEqual(int ar[], int n)
{
int sum = 0;
// Find the sum of first n-1
// elements of the array
for (int i = 0; i < n - 1; i++)
sum += ar[i];
// If sum equals to the last element
if (sum == ar[n - 1])
return true;
return false;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
if (isSumEqual(arr, 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
import java.io.*;
class GFG {
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
static boolean isSumEqual(int ar[], int n)
{
int sum = 0;
// Find the sum of first n-1
// elements of the array
for (int i = 0; i < n - 1; i++)
sum += ar[i];
// If sum equals to the last element
if (sum == ar[n - 1])
return true;
return false;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 10 };
int n = arr.length;
if (isSumEqual(arr, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by jit_t.
Python 3
# Python 3 implementation of the approach
# Function that returns true if sum of
# first n-1 elements of the array is
# equal to the last element
def isSumEqual(ar, n):
sum = 0
# Find the sum of first n-1
# elements of the array
for i in range(n - 1):
sum += ar[i]
# If sum equals to the last element
if (sum == ar[n - 1]):
return True
return False
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 10]
n = len(arr)
if (isSumEqual(arr, n)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the approach
using System;
class GFG {
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
static bool isSumEqual(int[] ar, int n)
{
int sum = 0;
// Find the sum of first n-1
// elements of the array
for (int i = 0; i < n - 1; i++)
sum += ar[i];
// If sum equals to the last element
if (sum == ar[n - 1])
return true;
return false;
}
// Driver code
static public void Main()
{
int[] arr = { 1, 2, 3, 4, 10 };
int n = arr.Length;
if (isSumEqual(arr, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by ajit
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 sum of
// first n-1 elements of the array is
// equal to the last element
function isSumEqual($ar, $n)
{
$sum = 0;
// Find the sum of first n-1
// elements of the array
for ($i = 0; $i < $n - 1; $i++)
$sum += $ar[$i];
// If sum equals to the last element
if ($sum == $ar[$n - 1])
return true;
return false;
}
// Driver code
$arr = array( 1, 2, 3, 4, 10 );
$n = count($arr);
if (isSumEqual($arr, $n))
echo "Yes";
else
echo "No";
// This code is contributed by AnkitRai01
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
function isSumEqual(ar, n)
{
let sum = 0;
// Find the sum of first n-1
// elements of the array
for (let i = 0; i < n - 1; i++)
sum += ar[i];
// If sum equals to the last element
if (sum == ar[n - 1])
return true;
return false;
}
let arr = [ 1, 2, 3, 4, 10 ];
let n = arr.length;
if (isSumEqual(arr, n))
document.write("Yes");
else
document.write("No");
</script>
Output:
Yes
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