使数组每个元素可被 4 整除的最小对求和运算
给定一个长度为n的正整数数组。我们的任务是找到转换数组的最小操作数，以使每个i的arr[i] % 4为零。在每个操作中，我们可以从数组中取出任意两个元素，将它们都删除，然后将它们的总和放回数组中。
例子：
Input : arr = {2 , 2 , 2 , 3 , 3}
Output : 3
Explanation: In 1 operation we pick 2 and 2 and put their sum back to the array , In 2 operation we pick 3 and 3 and do same for that ,now in 3 operation we pick 6 and 2 so overall 3 operation are required.
Input: arr = {4, 2, 2, 6, 6}
Output: 2
Explanation: In operation 1, we can take 2 and 2 and put back their sum i.e. 4. In operation 2, we can take 6 and 6 and put back their sum i.e. 12. And array becomes {4, 4, 12}.
方法：假设除以4时剩余1、2、3的元素计数为brr[1] 、 brr[2]和brr[3] 。
如果(brr[1] + 2 * brr[2] + 3 * brr[3])不是4的倍数，则不存在解。
现在贪婪地将brr[2]的元素与brr[2]和brr[1]的元素与brr[3] 配对。这有助于我们一次最多修复2 个元素。现在，我们要么只留下1 个brr[2]元素，要么没有。如果我们剩下1 个brr[2]元素，那么我们可以与剩余的2 个brr[1]或brr[3]元素配对。这将导致总共2 次操作。
最后，我们将只剩下brr[1]或brr[3]元素（如果可能的话）。这只能我们以一种方式解决。这需要其中的4 个，并在3 个操作中将它们全部固定在一起。因此，我们能够修复数组的所有元素。
下面是实现：
C++
// CPP program to find Minimum number
// of operations to convert an array
// so that arr[i] % 4 is zero.
#include
using namespace std;
// Function to find minimum operations.
int minimumOperations(int arr[], int n)
{
// Counting of all the elements
// leaving remainder 1, 2, 3 when
// divided by 4 in the array brr.
// at positions 1, 2 and 3 respectively.
int brr[] = { 0, 0, 0, 0 };
for (int i = 0; i < n; i++)
brr[arr[i] % 4]++;
// If it is possible to convert the
// array so that arr[i] % 4 is zero.
if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0)
{
// Pairing the elements of brr3 and brr1.
int min_opr = min(brr[3], brr[1]);
brr[3] -= min_opr;
brr[1] -= min_opr;
// Pairing the brr2 elements.
min_opr += brr[2] / 2;
// Assigning the remaining brr2 elements.
brr[2] %= 2;
// If we are left with one brr2 element.
if (brr[2]) {
// Here we need only two operations
// to convert the remaining one
// brr2 element to convert it.
min_opr += 2;
// Now there is no brr2 element.
brr[2] = 0;
// Remaining brr3 elements.
if (brr[3])
brr[3] -= 2;
// Remaining brr1 elements.
if (brr[1])
brr[1] -= 2;
}
// If we are left with brr1 and brr2
// elements then, we have to take four
// of them and fixing them all together
// in 3 operations.
if (brr[1])
min_opr += (brr[1] / 4) * 3;
if (brr[3])
min_opr += (brr[3] / 4) * 3;
// Returns the minimum operations.
return min_opr;
}
// If it is not possible to convert the array.
return -1;
}
// Driver function
int main()
{
int arr[] = { 1, 2, 3, 1, 2, 3, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << minimumOperations(arr, n);
} Java
// Java program to find Minimum number
// of operations to convert an array
// so that arr[i] % 4 is zero.
class GFG {
// Function to find minimum operations.
static int minimumOperations(int arr[], int n)
{
// Counting of all the elements
// leaving remainder 1, 2, 3 when
// divided by 4 in the array brr.
// at positions 1, 2 and 3 respectively.
int brr[] = { 0, 0, 0, 0 };
for (int i = 0; i < n; i++)
brr[arr[i] % 4]++;
// If it is possible to convert the
// array so that arr[i] % 4 is zero.
if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0)
{
// Pairing the elements of brr3 and brr1.
int min_opr = Math.min(brr[3], brr[1]);
brr[3] -= min_opr;
brr[1] -= min_opr;
// Pairing the brr2 elements.
min_opr += brr[2] / 2;
// Assigning the remaining brr2 elements.
brr[2] %= 2;
// If we are left with one brr2 element.
if (brr[2] == 1) {
// Here we need only two operations
// to convert the remaining one
// brr2 element to convert it.
min_opr += 2;
// Now there is no brr2 element.
brr[2] = 0;
// Remaining brr3 elements.
if (brr[3] == 1)
brr[3] -= 2;
// Remaining brr1 elements.
if (brr[1]== 1)
brr[1] -= 2;
}
// If we are left with brr1 and brr2
// elements then, we have to take four
// of them and fixing them all together
// in 3 operations.
if (brr[1] == 1)
min_opr += (brr[1] / 4) * 3;
if (brr[3] == 1)
min_opr += (brr[3] / 4) * 3;
// Returns the minimum operations.
return min_opr;
}
// If it is not possible to convert the array.
return -1;
}
// Driver function
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 1, 2, 3, 8 };
int n = arr.length;
System.out.println(minimumOperations(arr, n));
}
}
// This code is contributed by Prerna Saini.Python3
# Python program to
# find Minimum number
# of operations to
# convert an array
# so that arr[i] % 4 is zero.
# Function to find
# minimum operations.
def minimumOperations(arr,n):
# Counting of all the elements
# leaving remainder 1, 2, 3 when
# divided by 4 in the array brr.
# at positions 1, 2 and 3 respectively.
brr = [ 0, 0, 0, 0 ]
for i in range(n):
brr[arr[i] % 4]+=1;
# If it is possible to convert the
# array so that arr[i] % 4 is zero.
if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0):
# Pairing the elements
# of brr3 and brr1.
min_opr = min(brr[3], brr[1])
brr[3] -= min_opr
brr[1] -= min_opr
# Pairing the brr2 elements.
min_opr += brr[2] // 2
# Assigning the remaining
# brr2 elements.
brr[2] %= 2
# If we are left with
# one brr2 element.
if (brr[2]):
# Here we need only two operations
# to convert the remaining one
# brr2 element to convert it.
min_opr += 2
# Now there is no brr2 element.
brr[2] = 0
# Remaining brr3 elements.
if (brr[3]):
brr[3] -= 2
# Remaining brr1 elements.
if (brr[1]):
brr[1] -= 2
# If we are left with brr1 and brr2
# elements then, we have to take four
# of them and fixing them all together
# in 3 operations.
if (brr[1]):
min_opr += (brr[1] // 4) * 3
if (brr[3]):
min_opr += (brr[3] // 4) * 3
# Returns the minimum operations.
return min_opr
# If it is not possible to convert the array.
return -1
# Driver function
arr = [ 1, 2, 3, 1, 2, 3, 8 ]
n =len(arr)
print(minimumOperations(arr, n))
# This code is contributed
# by Anant Agarwal.C#
// C# program to find Minimum number
// of operations to convert an array
// so that arr[i] % 4 is zero.
using System;
class GFG {
// Function to find minimum operations.
static int minimumOperations(int []arr, int n)
{
// Counting of all the elements
// leaving remainder 1, 2, 3 when
// divided by 4 in the array brr.
// at positions 1, 2 and 3 respectively.
int []brr = { 0, 0, 0, 0 };
for (int i = 0; i < n; i++)
brr[arr[i] % 4]++;
// If it is possible to convert the
// array so that arr[i] % 4 is zero.
if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0)
{
// Pairing the elements of brr3 and brr1.
int min_opr = Math.Min(brr[3], brr[1]);
brr[3] -= min_opr;
brr[1] -= min_opr;
// Pairing the brr2 elements.
min_opr += brr[2] / 2;
// Assigning the remaining brr2 elements.
brr[2] %= 2;
// If we are left with one brr2 element.
if (brr[2] == 1) {
// Here we need only two operations
// to convert the remaining one
// brr2 element to convert it.
min_opr += 2;
// Now there is no brr2 element.
brr[2] = 0;
// Remaining brr3 elements.
if (brr[3] == 1)
brr[3] -= 2;
// Remaining brr1 elements.
if (brr[1]== 1)
brr[1] -= 2;
}
// If we are left with brr1 and brr2
// elements then, we have to take four
// of them and fixing them all together
// in 3 operations.
if (brr[1] == 1)
min_opr += (brr[1] / 4) * 3;
if (brr[3] == 1)
min_opr += (brr[3] / 4) * 3;
// Returns the minimum operations.
return min_opr;
}
// If it is not possible to convert the array.
return -1;
}
// Driver function
public static void Main()
{
int []arr = { 1, 2, 3, 1, 2, 3, 8 };
int n = arr.Length;
Console.WriteLine(minimumOperations(arr, n));
}
}
// This code is contributed by vt_mPHP
Javascript
输出：
3