检查给定数组的和是否可以通过将数组元素减少 K 来减少到 0
原文:https://www . geeksforgeeks . org/check-if-sum-给定数组可以通过将数组元素减少 k 来减少到 0/
给定一个由 N 个整数和一个整数 K、组成的数组arr【】】，任务是检查数组的和是否可以通过将数组元素减去 K 任意次数而减少为 0。
示例:
输入: arr[ ]= {-3，2，-1，5，1}，K=2 输出:【是】 说明: 数组之和为 4。因此，将任意索引处的两个元素减 K( = 2)，使数组的和为 0。 输入: arr[ ]= {1，-6，2，2}，K=1 输出:“否”
方法:按照以下步骤解决问题:
- 遍历数组并计算给定数组的和。
- 根据和的值，出现以下情况:
- 如果总和= 0: 不需要操作。因此，答案是“是的”。
- If sum > 0: Sum 只有在 sum 是 K 的倍数时，才能化简为 0 。如果和不是 K 的倍数，则打印“否”。否则，打印“是”。
- 如果求和< 0: 只需打印“否”。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the
// sum can be made 0 or not
int sumzero(int arr[], int N, int K)
{
// Stores sum of array elements
int sum = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
sum += arr[i];
}
if (sum == 0)
cout << "Yes";
else if (sum > 0) {
if (sum % K == 0)
cout << "Yes";
else
cout << "No";
}
else
cout << "No";
return 0;
}
// Driver Code
int main()
{
int K, N;
// Given array arr[]
int arr1[] = { 1, -6, 2, 2 };
K = 1;
N = sizeof(arr1) / sizeof(arr1[0]);
sumzero(arr1, N, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if the
// sum can be made 0 or not
static int sumzero(int arr[], int N, int K)
{
// Stores sum of array elements
int sum = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
sum += arr[i];
}
if (sum == 0)
System.out.print("Yes");
else if (sum > 0) {
if (sum % K == 0)
System.out.print("Yes");
else
System.out.print("No");
}
else
System.out.print("No");
return 0;
}
// Driver Code
public static void main(String[] args)
{
int K, N;
// Given array arr[]
int arr1[] = { 1, -6, 2, 2 };
K = 1;
N = arr1.length;
sumzero(arr1, N, K);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program for the above approach
# Function to check if the
# sum can be made 0 or not
def sumzero(arr, N, K) :
# Stores sum of array elements
sum = 0;
# Traverse the array
for i in range(N) :
sum += arr[i];
if (sum == 0) :
print("Yes");
elif (sum > 0) :
if (sum % K == 0) :
print("Yes");
else :
print("No");
else :
print("No");
# Driver Code
if __name__ == "__main__" :
# Given array arr[]
arr1 = [ 1, -6, 2, 2 ];
K = 1;
N = len(arr1);
sumzero(arr1, N, K);
# This code is contributed by AnkThon
C
// C# program for the above approach
using System;
class GFG{
// Function to check if the
// sum can be made 0 or not
static int sumzero(int []arr, int N, int K)
{
// Stores sum of array elements
int sum = 0;
// Traverse the array
for (int i = 0; i < N; i++)
{
sum += arr[i];
}
if (sum == 0)
Console.Write("Yes");
else if (sum > 0)
{
if (sum % K == 0)
Console.Write("Yes");
else
Console.Write("No");
}
else
Console.Write("No");
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int K, N;
// Given array []arr
int []arr1 = { 1, -6, 2, 2 };
K = 1;
N = arr1.Length;
sumzero(arr1, N, K);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript program for the above approach
// Function to check if the
// sum can be made 0 or not
function sumzero(arr , N , K)
{
// Stores sum of array elements
var sum = 0;
// Traverse the array
for (i = 0; i < N; i++) {
sum += arr[i];
}
if (sum == 0)
document.write("Yes");
else if (sum > 0) {
if (sum % K == 0)
document.write("Yes");
else
document.write("No");
}
else
document.write("No");
return 0;
}
// Driver Code
var K, N;
// Given array arr
var arr1 = [ 1, -6, 2, 2 ];
K = 1;
N = arr1.length;
sumzero(arr1, N, K);
// This code contributed by gauravrajput1
</script>
Output:
No
时间复杂度:O(N) T5辅助空间:** O(1)
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