给定大小为N的整数数组arr []和整数K ，任务是找到执行给定操作后将最后减少为零的索引。操作说明如下：
- 从arr [0]到arr [N – 1] ，将每个元素更新为arr [i] = arr [i] – K。
- 如果ARR [I]
- 重复上述步骤，直到所有元素都减小为0为止。
打印将成为最后一个变为零的索引。
例子：
Input: arr[] = { 3, 2, 5, 7, 2, 9 }, K = 4
Output: 5
Operation 1: arr[] = {0, 0, 1, 3, 0, 5}
Operation 2: arr[] = {0, 0, 0, 0, 0, 1}
Operation 3: arr[] = {0, 0, 0, 0, 0, 0}
Index 5 is the last to reduce.
Input: arr[] = { 31, 12, 25, 27, 32, 19 }, K = 5
Output: 4
方法：在每个步骤中，将特定索引处的元素减去K。因此，特定元素需要ceil(arr [i] / K)或(arr [i] + K – 1)/ K步减少为零。因此，所需的索引由具有最大(arr [i] + K – 1)/ K值的数组索引给出。如果存在的最大值不止一次，则在从0到N – 1的操作中返回最大的索引。
下面是上述方法的实现：
CPP
// C++ implementation of the approach
#include
using namespace std;
// Function that returns the index
// which will be the last to become
// zero after performing given operation
int findIndex(int a[], int n, int k)
{
// Initialize the result
int index = -1, max_ceil = INT_MIN;
for (int i = 0; i < n; i++) {
// Finding the ceil value
// of each index
a[i] = (a[i] + k - 1) / k;
}
for (int i = 0; i < n; i++) {
// Finding the index with
// maximum ceil value
if (a[i] >= max_ceil) {
max_ceil = a[i];
index = i;
}
}
return index;
}
// Driver code
int main()
{
int arr[] = { 31, 12, 25, 27, 32, 19 };
int K = 5;
int N = sizeof(arr) / sizeof(arr[0]);
cout << findIndex(arr, N, K);
return 0;
} Java
// Java implementation of the approach
import java .io.*;
class GFG
{
// Function that returns the index
// which will be the last to become
// zero after performing given operation
static int findIndex(int[] a, int n, int k)
{
// Initialize the result
int index = -1, max_ceil = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
{
// Finding the ceil value
// of each index
a[i] = (a[i] + k - 1) / k;
}
for (int i = 0; i < n; i++)
{
// Finding the index with
// maximum ceil value
if (a[i] >= max_ceil)
{
max_ceil = a[i];
index = i;
}
}
return index;
}
// Driver code
static public void main (String[] args)
{
int []arr = { 31, 12, 25, 27, 32, 19 };
int K = 5;
int N = arr.length ;
System.out.print(findIndex(arr, N, K));
}
}
// This code is contributed by anuj_67..Python
# Python implementation of the approach
# Function that returns the index
# which will be the last to become
# zero after performing given operation
def findIndex(a, n, k):
# Initialize the result
index = -1
max_ceil = -10**9
for i in range(n):
# Finding the ceil value
# of each index
a[i] = (a[i] + k - 1) // k
for i in range(n):
# Finding the index with
# maximum ceil value
if (a[i] >= max_ceil):
max_ceil = a[i]
index = i
return index
# Driver code
arr = [31, 12, 25, 27, 32, 19]
K = 5
N = len(arr)
print(findIndex(arr, N, K))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns the index
// which will be the last to become
// zero after performing given operation
static int findIndex(int[] a, int n, int k)
{
// Initialize the result
int index = -1, max_ceil = int.MinValue;
for (int i = 0; i < n; i++)
{
// Finding the ceil value
// of each index
a[i] = (a[i] + k - 1) / k;
}
for (int i = 0; i < n; i++)
{
// Finding the index with
// maximum ceil value
if (a[i] >= max_ceil)
{
max_ceil = a[i];
index = i;
}
}
return index;
}
// Driver code
static public void Main ()
{
int []arr = { 31, 12, 25, 27, 32, 19 };
int K = 5;
int N = arr.Length ;
Console.WriteLine(findIndex(arr, N, K));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
4