生成一个 N 长度阵列，每个子阵列的和可被 K 整除
原文:https://www . geeksforgeeks . org/generate-an-n-length-array-having-sum-by-k-除尽/
给定两个正整数 N 和 K ，任务是生成一个由 N 个不同整数组成的数组，使得构造的数组的每个子数组的元素之和可被 K 整除。
示例:
输入: N = 3，K = 3 输出: 3 6 9 解释: 合成阵列的子阵列为{3}、{6}、{3，6}、{9}、{6，9}、{3，6，9}。所有这些子阵的和都可以被 k 整除
输入: N = 5，K = 1 T3】输出: 1 2 3 4 5
方法:按照以下步骤解决问题:
- 由于每个子阵列的元素之和需要被 K 整除，所以最理想的方法是构建一个数组，其中每个元素都是 K 的倍数。
- 因此，迭代一个从 i = 1 到 i = N 的循环，对于 i 的每个值，打印 K * i 。
下面是上述方法的实现:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct an array
// with sum of each subarray
// divisible by K
void construct_Array(int N, int K)
{
// Traverse a loop from 1 to N
for (int i = 1; i <= N; i++) {
// Print i-th multiple of K
cout << K * i << " ";
}
}
// Driver Code
int main()
{
int N = 3, K = 3;
construct_Array(N, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to construct an array
// with sum of each subarray
// divisible by K
static void construct_Array(int N, int K)
{
// Traverse a loop from 1 to N
for (int i = 1; i <= N; i++)
{
// Print i-th multiple of K
System.out.print(K * i + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 3, K = 3;
construct_Array(N, K);
}
}
// This code is contributed by code hunt.
Python 3
# Python program for the above approach
# Function to construct an array
# with sum of each subarray
# divisible by K
def construct_Array(N, K) :
# Traverse a loop from 1 to N
for i in range(1, N + 1):
# Pri-th multiple of K
print(K * i, end = " ")
# Driver Code
N = 3
K = 3
construct_Array(N, K)
# This code is contributed by splevel62.
C
// C# program to implement
// the above approach
using System;
public class GFG
{
// Function to construct an array
// with sum of each subarray
// divisible by K
static void construct_Array(int N, int K)
{
// Traverse a loop from 1 to N
for (int i = 1; i <= N; i++)
{
// Print i-th multiple of K
Console.Write(K * i + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 3, K = 3;
construct_Array(N, K);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript program for the above approach
// Function to construct an array
// with sum of each subarray
// divisible by K
function construct_Array(N, K)
{
// Traverse a loop from 1 to N
for (let i = 1; i <= N; i++) {
// Print i-th multiple of K
document.write( K * i + " ");
}
}
// Driver Code
let N = 3, K = 3;
construct_Array(N, K);
// This code is contributed by todaysgaurav
</script>
Output:
3 6 9
时间复杂度:O(N) T5辅助空间:** O(1)
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