通过在每第 I 步从一个数组元素中减去 K^i 使所有数组元素等于 0 来修改数组
原文:https://www . geeksforgeeks . org/通过将所有数组元素从每 I 步数组元素中减去 ki 等于 0 来修改数组/
给定一个大小为 N、的数组 arr[] ，任务是在 i th 步骤中，通过从数组元素中减去 K i 来检查是否有可能将所有数组元素转换为 0 s。如果可以，则打印“是”。否则，打印“否”。
示例:
输入: N = 5，K = 2，arr[] = {8，0，3，4，80} 输出:是 解释: 一种可能的操作顺序如下:
- 从 arr2中减去 2 0 。此后，数组修改为，arr[] = {8，0，2，4，32}。
- 从 arr2中减去 2 1 。此后，数组修改为，arr[] = {8，0，0，4，32}。
- 从 arr3中减去 2 2 。此后，数组修改为，arr[] = {8，0，0，0，32}。
- 从 arr1中减去 2 3 。此后，数组修改为，arr[] = {0，0，0，0，32}。
- 不要从任何数组元素中减去 2 4 。
- 从 arr4中减去 2 5 。此后，数组修改为，arr[] = {0，0，0，0，0}。
输入: N = 3，K = 2，arr[] = {0，1，3 } T3】输出:否
方法:按照以下步骤解决问题:
- 初始化一个向量，比如 V、来存储 K 所有可能的力量。
- 另外，初始化一个映射< int，int > ，比如 MP ，来存储是否使用了 K 的电源。
- 初始化一个变量，说 X 为 1，存储 K. 的幂的计数
- 迭代直到 X 小于T5INT _ MAX并执行以下步骤:
- 将 X 的值推入向量。
- 将乘以乘以 K.
- 使用变量 i 迭代范围【0，N–1】，并执行以下步骤:
- 如果 i 小于 N，则打印“否”，作为数组元素不能做成 0 。否则，打印“是”。
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 whether all array
// elements can be made zero or not
string isMakeZero(int arr[], int N, int K)
{
// Stores if a power of K has
// already been subtracted or not
map<int, int> MP;
// Stores all the Kth power
vector<int> V;
int X = 1;
int i;
// Iterate until X is
// less than INT_MAX
while (X > 0 && X < INT_MAX) {
V.push_back(X);
X *= K;
}
// Traverse the array arr[]
for (i = 0; i < N; i++) {
// Iterate over the range [0, M]
for (int j = V.size() - 1; j >= 0; j--) {
// If MP[V[j]] is 0 and V[j]
// is less than or equal to arr[i]
if (MP[V[j]] == 0 && V[j] <= arr[i]) {
arr[i] -= V[j];
MP[V[j]] = 1;
}
}
// If arr[i] is not 0
if (arr[i] != 0)
break;
}
// If i is less than N
if (i < N)
return "No";
// Otherwise,
else
return "Yes";
}
// Driver code
int main()
{
int arr[] = { 8, 0, 3, 4, 80 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout << isMakeZero(arr, 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.ArrayList;
import java.util.HashMap;
class GFG {
// Function to check whether all array
// elements can be made zero or not
static String isMakeZero(int arr[], int N, int K)
{
// Stores if a power of K has
// already been subtracted or not
HashMap<Integer, Integer> MP = new HashMap<Integer, Integer>();
// Stores all the Kth power
ArrayList<Integer> V = new ArrayList<Integer>();
int X = 1;
int i;
// Iterate until X is
// less than INT_MAX
while (X > 0 && X < Integer.MAX_VALUE) {
V.add(X);
X *= K;
}
// Traverse the array arr[]
for (i = 0; i < N; i++) {
// Iterate over the range [0, M]
for (int j = V.size() - 1; j >= 0; j--) {
// If MP[V[j]] is 0 and V[j]
// is less than or equal to arr[i]
if (MP.containsKey(V.get(j)) == false && V.get(j) <= arr[i]) {
arr[i] -= V.get(j);
MP.put(V.get(j), 1);
}
}
// If arr[i] is not 0
if (arr[i] != 0)
break;
}
// If i is less than N
if (i < N)
return "No";
// Otherwise,
else
return "Yes";
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 0, 3, 4, 80 };
int N = arr.length;
int K = 2;
System.out.println(isMakeZero(arr, N, K));
}
}
// This code is contributed by _saurabh_jaiswal.
Python 3
# Python Program for the above approach
# Function to check whether all array
# elements can be made zero or not
def isMakeZero(arr, N, K):
# Stores if a power of K has
# already been subtracted or not
MP = {}
# Stores all the Kth power
V = []
X = 1
# Iterate until X is
# less than INT_MAX
while (X > 0 and X < 10**20):
V.append(X)
X *= K
# Traverse the array arr[]
for i in range(0, N, 1):
# Iterate over the range [0, M]
for j in range(len(V) - 1, -1, -1):
# If MP[V[j]] is 0 and V[j]
# is less than or equal to arr[i]
if (V[j] not in MP and V[j] <= arr[i]):
arr[i] -= V[j]
MP[V[j]] = 1
# If arr[i] is not 0
if (arr[i] != 0):
break
# If i is less than N - 1
if (i < N - 1):
return "No"
# Otherwise,
else:
return "Yes"
# Driver code
arr = [8, 0, 3, 4, 80]
N = len(arr)
K = 2
print(isMakeZero(arr, N, K))
# This code is contributed by _saurabh_jaiswal
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check whether all array
// elements can be made zero or not
static string isMakeZero(int[] arr, int N, int K)
{
// Stores if a power of K has
// already been subtracted or not
Dictionary<int,int> MP = new Dictionary<int,int>();
// Stores all the Kth power
List<int> V = new List<int>();
int X = 1;
int i;
// Iterate until X is
// less than INT_MAX
while (X > 0 && X < Int32.MaxValue) {
V.Add(X);
X *= K;
}
// Traverse the array arr[]
for (i = 0; i < N; i++) {
// Iterate over the range [0, M]
for (int j = V.Count - 1; j >= 0; j--) {
// If MP[V[j]] is 0 and V[j]
// is less than or equal to arr[i]
if (MP.ContainsKey(V[j]) == false && V[j] <= arr[i]) {
arr[i] -= V[j];
MP[V[j]] = 1;
}
}
// If arr[i] is not 0
if (arr[i] != 0)
break;
}
// If i is less than N
if (i < N)
return "No";
// Otherwise,
else
return "Yes";
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 8, 0, 3, 4, 80 };
int N = arr.Length;
int K = 2;
Console.WriteLine(isMakeZero(arr, N, K));
}
}
// This code is contributed by splevel62.
java 描述语言
<script>
// JavaScript Program for the above approach
// Function to check whether all array
// elements can be made zero or not
function isMakeZero(arr, N, K) {
// Stores if a power of K has
// already been subtracted or not
let MP = new Map();
// Stores all the Kth power
let V = [];
let X = 1;
let i;
// Iterate until X is
// less than INT_MAX
while (X > 0 && X < Number.MAX_VALUE) {
V.push(X);
X *= K;
}
// Traverse the array arr[]
for (i = 0; i < N; i++) {
// Iterate over the range [0, M]
for (let j = V.length - 1; j >= 0; j--) {
// If MP[V[j]] is 0 and V[j]
// is less than or equal to arr[i]
if (MP.has(V[j]) == false && V[j] <= arr[i]) {
arr[i] -= V[j];
MP.set(V[j], 1);
}
}
// If arr[i] is not 0
if (arr[i] != 0)
break;
}
// If i is less than N
if (i < N)
return "No";
// Otherwise,
else
return "Yes";
}
// Driver code
let arr = [8, 0, 3, 4, 80];
let N = arr.length;
let K = 2;
document.write(isMakeZero(arr, N, K));
// This code is contributed by Potta Lokesh
</script>
Output:
Yes
时间复杂度:O(N * logK(INT _ MAX)) 辅助空间:O(logK(INT _ MAX))
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