通过 k 次运算的最小增量以使所有元素相等 · GeeksForGeeks 中文系列教程

# 通过 k 次运算的最小增量以使所有元素相等 > 原文： [https://www.geeksforgeeks.org/minimum-increment-k-operations-make-elements-equal/](https://www.geeksforgeeks.org/minimum-increment-k-operations-make-elements-equal/) 给您一个 n 元素数组，您必须找到使数组的所有元素相等所需的操作数。单个操作可以将元素增加 k。如果不可能使所有元素相等，则打印-1。 **示例**： ``` Input : arr[] = {4, 7, 19, 16}, k = 3 Output : 10 Input : arr[] = {4, 4, 4, 4}, k = 3 Output : 0 Input : arr[] = {4, 2, 6, 8}, k = 3 Output : -1 ``` 为了解决这个问题，我们需要检查所有元素是否可以相等，也只能通过从元素值中增加 k 来检查。为此，我们必须检查任何两个元素的差应始终被 k 整除。如果是这样，那么所有元素都可以变得相等，否则无论如何都不能通过使它们递增 k 来变得相等。同样，可以通过找到所有元素的（max – Ai）/ k 值来计算所需的操作次数。其中 max 是数组的最大元素。算法： ``` // iterate for all elements for (int i=0; i<n; i++) { // check if element can make equal to max // or not if not then return -1 if ((max - arr[i]) % k != 0 ) return -1; // else update res for required operations else res += (max - arr[i]) / k ; } return res; ``` ## C++ ```cpp // Program to make all array equal #include <bits/stdc++.h> using namespace std; // function for calculating min operations int minOps(int arr[], int n, int k) { // max elements of array int max = *max_element(arr, arr + n); int res = 0; // iterate for all elements for (int i = 0; i < n; i++) { // check if element can make equal to // max or not if not then return -1 if ((max - arr[i]) % k != 0) return -1; // else update res for required operations else res += (max - arr[i]) / k; } // return result return res; } // driver program int main() { int arr[] = { 21, 33, 9, 45, 63 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 6; cout << minOps(arr, n, k); return 0; } ``` ## Java ```java // Program to make all array equal import java.io.*; import java.util.Arrays; class GFG { // function for calculating min operations static int minOps(int arr[], int n, int k) { // max elements of array Arrays.sort(arr); int max = arr[arr.length - 1]; int res = 0; // iterate for all elements for (int i = 0; i < n; i++) { // check if element can make equal to // max or not if not then return -1 if ((max - arr[i]) % k != 0) return -1; // else update res for required operations else res += (max - arr[i]) / k; } // return result return res; } // Driver program public static void main(String[] args) { int arr[] = { 21, 33, 9, 45, 63 }; int n = arr.length; int k = 6; System.out.println(minOps(arr, n, k)); } } // This code is contributed by vt_m ``` ## Python3 ```py # Python3 Program to make all array equal # function for calculating min operations def minOps(arr, n, k): # max elements of array max1 = max(arr) res = 0 # iterate for all elements for i in range(0, n): # check if element can make equal to # max or not if not then return -1 if ((max1 - arr[i]) % k != 0): return -1 # else update res fo # required operations else: res += (max1 - arr[i]) / k # return result return int(res) # driver program arr = [21, 33, 9, 45, 63] n = len(arr) k = 6 print(minOps(arr, n, k)) # This code is contributed by # Smitha Dinesh Semwal ``` ## C# ```cs // C# program Minimum increment by // k operations to make all elements equal. using System; class GFG { // function for calculating min operations static int minOps(int[] arr, int n, int k) { // max elements of array Array.Sort(arr); int max = arr[arr.Length - 1]; int res = 0; // iterate for all elements for (int i = 0; i < n; i++) { // check if element can make // equal to max or not if not // then return -1 if ((max - arr[i]) % k != 0) return -1; // else update res for required // operations else res += (max - arr[i]) / k; } // return result return res; } // Driver program public static void Main() { int[] arr = { 21, 33, 9, 45, 63 }; int n = arr.Length; int k = 6; Console.Write(minOps(arr, n, k)); } } // This code is contributed by nitin mittal. ``` ## PHP ```php <?php // Program to make all array equal // function for calculating // min operations function minOps($arr, $n, $k) { // max elements of array $max = max($arr); $res = 0; // iterate for all elements for ($i = 0; $i < $n; $i++) { // check if element can // make equal to max or // not if not then return -1 if (($max - $arr[$i]) % $k != 0) return -1; // else update res for // required operations else $res += ($max - $arr[$i]) / $k; } // return result return $res; } // Driver Code $arr = array(21, 33, 9, 45, 63); $n = count($arr); $k = 6; print (minOps($arr, $n, $k)); // This code is contributed // by Manish Shaw(manishshaw1) ?> ``` **输出**： ``` 24 ``` * * * * * *