给定大小为N的整数arr []整数K ，任务是通过最多执行K次以下操作来找到最大化数组的第一个元素：
- 选择一对索引i和j (0≤i，j≤N-1) ，使得| i − j | = 1且arr i > 0 。
- 在这两个索引上分别设置arr i = arr i − 1和arr j = arr j +1 。
例子：
Input: arr[ ] = {1, 0, 3, 2}, K = 5
Output: 3
Explanation:
One of the possible set of operations can be:
Operation 1: Select i = 3 and j = 2. Therefore, the array modifies to {1, 1, 2, 2}.
Operation 2: Select i = 3 and j = 2. Therefore, the array modifies to {1, 2, 1, 2}.
Operation 3: Select i = 2 and j = 1. Therefore, the array modifies to {2, 1, 1, 2}.
Operation 4: Select i = 2 and j = 1. Therefore, the array modifies to {3, 0, 1, 2}.
Input: arr[] = {5, 1}, K = 2
Output: 6
方法：请按照以下步骤解决问题：
- 在任何时候，最佳选择索引i和j最接近数组的第一个元素，且i> j 。
- 因此，对于每个操作，请从左到右遍历数组，并将元素移近第一个元素。
- 如果所有元素都在某个位置处在第一位置，则停止遍历并打印第一个数组元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to maximize
// the first array element
int getMax(int arr[], int N, int K)
{
// Traverse the array
for (int i = 1; i < N; i++) {
// Initialize cur_val to a[i]
int cur_val = arr[i];
// If all operations
// are not over yet
while (K >= i) {
// If current value is
// greater than zero
if (cur_val > 0) {
// Incrementing first
// element of array by 1
arr[0] = arr[0] + 1;
// Decrementing current
// value of array by 1
cur_val = cur_val - 1;
// Decrementing number
// of operations by i
K = K - i;
}
// If current value is
// zero, then break
else
break;
}
}
// Print first array element
cout << arr[0];
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 0, 3, 2 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Given K
int K = 5;
// Prints the maximum
// possible value of the
// first array element
getMax(arr, N, K);
return 0;
} Java
// Java program for the above approach
class GFG{
// Function to maximize
// the first array element
static void getMax(int arr[], int N, int K)
{
// Traverse the array
for (int i = 1; i < N; i++)
{
// Initialize cur_val to a[i]
int cur_val = arr[i];
// If all operations
// are not over yet
while (K >= i)
{
// If current value is
// greater than zero
if (cur_val > 0)
{
// Incrementing first
// element of array by 1
arr[0] = arr[0] + 1;
// Decrementing current
// value of array by 1
cur_val = cur_val - 1;
// Decrementing number
// of operations by i
K = K - i;
}
// If current value is
// zero, then break
else
break;
}
}
// Print first array element
System.out.print(arr[0]);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 0, 3, 2 };
// Size of the array
int N = arr.length;
// Given K
int K = 5;
// Prints the maximum
// possible value of the
// first array element
getMax(arr, N, K);
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 program for the above approach
# Function to maximize
# the first array element
def getMax(arr, N, K):
# Traverse the array
for i in range(1, N, 1):
# Initialize cur_val to a[i]
cur_val = arr[i]
# If all operations
# are not over yet
while (K >= i):
# If current value is
# greater than zero
if (cur_val > 0):
# Incrementing first
# element of array by 1
arr[0] = arr[0] + 1
# Decrementing current
# value of array by 1
cur_val = cur_val - 1
# Decrementing number
# of operations by i
K = K - i
# If current value is
# zero, then break
else:
break
# Print first array element
print(arr[0])
# Driver Code
if __name__ == '__main__':
# Given array
arr = [ 1, 0, 3, 2 ]
# Size of the array
N = len(arr)
# Given K
K = 5
# Prints the maximum
# possible value of the
# first array element
getMax(arr, N, K)
# This code is contributed by SURENDRA_GANGWARC#
// C# program for the above approach
using System;
class GFG{
// Function to maximize
// the first array element
static void getMax(int[] arr, int N,
int K)
{
// Traverse the array
for(int i = 1; i < N; i++)
{
// Initialize cur_val to a[i]
int cur_val = arr[i];
// If all operations
// are not over yet
while (K >= i)
{
// If current value is
// greater than zero
if (cur_val > 0)
{
// Incrementing first
// element of array by 1
arr[0] = arr[0] + 1;
// Decrementing current
// value of array by 1
cur_val = cur_val - 1;
// Decrementing number
// of operations by i
K = K - i;
}
// If current value is
// zero, then break
else
break;
}
}
// Print first array element
Console.Write(arr[0]);
}
// Driver code
static void Main()
{
// Given array
int[] arr = { 1, 0, 3, 2 };
// Size of the array
int N = arr.Length;
// Given K
int K = 5;
// Prints the maximum
// possible value of the
// first array element
getMax(arr, N, K);
}
}
// This code is contributed by divyesh072019输出：
3时间复杂度： O(N)
辅助空间： O(1)