给定一个由n个整数和一个整数K组成的数组arr [] 。任务是找到从数组中选择一个或多个元素的方法，以使所选整数的平均值等于给定数K。
例子：
Input: arr[] = {7, 9, 8, 9}, K = 8
Output: 5
{8}, {7, 9}, {7, 9}, {7, 8, 9} and {7, 8, 9}
Input: arr[] = {3, 6, 2, 8, 7, 6, 5, 9}, K = 5
Output: 19
Input: arr[] = {6, 6, 9}, K = 8
Output: 0
简单方法：一个简单的解决方案是尝试所有可能性，因为N可以很大。时间复杂度可以是2 N。
高效方法：可以通过使用动态编程来解决上述问题，从而优化上述方法。假设我们在第i_个索引处，令val为该索引的当前值。我们有两种可能性可以选择答案中的该元素，也可以放弃该元素。因此，我们现在完成了。我们还将跟踪当前选定元素集中的元素数量。
以下是递归公式。
ways(index, sum, count)
= ways(index - 1, sum, count)
+ ways(index - 1, sum + arr[index], count + 1)
下面是上述方法的实现：
C++
#include
using namespace std;
#define MAX_INDEX 51
#define MAX_SUM 2505
// This dp array is used to store our values
// so that we don't have to calculate same
// values again and again
int dp[MAX_INDEX][MAX_SUM][MAX_INDEX];
int waysutil(int index, int sum, int count,
vector& arr, int K)
{
// Base cases
// Index can't be less than 0
if (index < 0)
return 0;
if (index == 0) {
// No element is picked hence
// average cannot be calculated
if (count == 0)
return 0;
int remainder = sum % count;
// If remainder is non zero, we cannot
// divide the sum by count i.e. the average
// will not be an integer
if (remainder != 0)
return 0;
int average = sum / count;
// If we find an average return 1
if (average == K)
return 1;
}
// If we have already calculated this function
// simply return it instead of calculating it again
if (dp[index][sum][count] != -1)
return dp[index][sum][count];
// If we don't pick the current element
// simple recur for index -1
int dontpick = waysutil(index - 1,
sum, count, arr, K);
// If we pick the current element add it to
// our current sum and increment count by 1
int pick = waysutil(index - 1,
sum + arr[index],
count + 1, arr, K);
int total = pick + dontpick;
// Store the value for the current function
dp[index][sum][count] = total;
return total;
}
// Function to return the number of ways
int ways(int N, int K, int* arr)
{
vector Arr;
// Push -1 at the beginning to
// make it 1-based indexing
Arr.push_back(-1);
for (int i = 0; i < N; ++i) {
Arr.push_back(arr[i]);
}
// Initialize dp array by -1
memset(dp, -1, sizeof dp);
// Call recursive function
// waysutil to calculate total ways
int answer = waysutil(N, 0, 0, Arr, K);
return answer;
}
// Driver code
int main()
{
int arr[] = { 3, 6, 2, 8, 7, 6, 5, 9 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 5;
cout << ways(N, K, arr);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
static int MAX_INDEX = 51;
static int MAX_SUM = 2505;
// This dp array is used to store our values
// so that we don't have to calculate same
// values again and again
static int[][][] dp = new int[MAX_INDEX][MAX_SUM][MAX_INDEX];
static int waysutil(int index, int sum, int count,
Vector arr, int K)
{
// Base cases
// Index can't be less than 0
if (index < 0)
{
return 0;
}
if (index == 0)
{
// No element is picked hence
// average cannot be calculated
if (count == 0)
{
return 0;
}
int remainder = sum % count;
// If remainder is non zero, we cannot
// divide the sum by count i.e. the average
// will not be an integer
if (remainder != 0)
{
return 0;
}
int average = sum / count;
// If we find an average return 1
if (average == K)
{
return 1;
}
}
// If we have already calculated this function
// simply return it instead of calculating it again
if (dp[index][sum][count] != -1)
{
return dp[index][sum][count];
}
// If we don't pick the current element
// simple recur for index -1
int dontpick = waysutil(index - 1,
sum, count, arr, K);
// If we pick the current element add it to
// our current sum and increment count by 1
int pick = waysutil(index - 1,
sum + arr.get(index),
count + 1, arr, K);
int total = pick + dontpick;
// Store the value for the current function
dp[index][sum][count] = total;
return total;
}
// Function to return the number of ways
static int ways(int N, int K, int[] arr)
{
Vector Arr = new Vector<>();
// Push -1 at the beginning to
// make it 1-based indexing
Arr.add(-1);
for (int i = 0; i < N; ++i)
{
Arr.add(arr[i]);
}
// Initialize dp array by -1
for (int i = 0; i < MAX_INDEX; i++)
{
for (int j = 0; j < MAX_SUM; j++)
{
for (int l = 0; l < MAX_INDEX; l++)
{
dp[i][j][l] = -1;
}
}
}
// Call recursive function
// waysutil to calculate total ways
int answer = waysutil(N, 0, 0, Arr, K);
return answer;
}
// Driver code
public static void main(String args[])
{
int arr[] = {3, 6, 2, 8, 7, 6, 5, 9};
int N = arr.length;
int K = 5;
System.out.println(ways(N, K, arr));
}
}
/* This code contributed by PrinciRaj1992 */ Python3
# Python implementation of above approach
import numpy as np
MAX_INDEX = 51
MAX_SUM = 2505
# This dp array is used to store our values
# so that we don't have to calculate same
# values again and again
# Initialize dp array by -1
dp = np.ones((MAX_INDEX,MAX_SUM,MAX_INDEX)) * -1;
def waysutil(index, sum, count, arr, K) :
# Base cases
# Index can't be less than 0
if (index < 0) :
return 0;
if (index == 0) :
# No element is picked hence
# average cannot be calculated
if (count == 0) :
return 0;
remainder = sum % count;
# If remainder is non zero, we cannot
# divide the sum by count i.e. the average
# will not be an integer
if (remainder != 0) :
return 0;
average = sum // count;
# If we find an average return 1
if (average == K) :
return 1;
# If we have already calculated this function
# simply return it instead of calculating it again
if (dp[index][sum][count] != -1) :
return dp[index][sum][count];
# If we don't pick the current element
# simple recur for index -1
dontpick = waysutil(index - 1,
sum, count, arr, K);
# If we pick the current element add it to
# our current sum and increment count by 1
pick = waysutil(index - 1,
sum + arr[index],
count + 1, arr, K);
total = pick + dontpick;
# Store the value for the current function
dp[index][sum][count] = total;
return total;
# Function to return the number of ways
def ways(N, K, arr) :
Arr = [];
# Push -1 at the beginning to
# make it 1-based indexing
Arr.append(-1);
for i in range(N) :
Arr.append(arr[i]);
# Call recursive function
# waysutil to calculate total ways
answer = waysutil(N, 0, 0, Arr, K);
return answer;
# Driver code
if __name__ == "__main__" :
arr = [ 3, 6, 2, 8, 7, 6, 5, 9 ];
N =len(arr);
K = 5;
print(ways(N, K, arr));
# This code is contributed by AnkitRai01C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
static int MAX_INDEX = 51;
static int MAX_SUM = 2505;
// This dp array is used to store our values
// so that we don't have to calculate same
// values again and again
static int[,,] dp = new int[MAX_INDEX, MAX_SUM, MAX_INDEX];
static int waysutil(int index, int sum, int count,
List arr, int K)
{
// Base cases
// Index can't be less than 0
if (index < 0)
{
return 0;
}
if (index == 0)
{
// No element is picked hence
// average cannot be calculated
if (count == 0)
{
return 0;
}
int remainder = sum % count;
// If remainder is non zero, we cannot
// divide the sum by count i.e. the average
// will not be an integer
if (remainder != 0)
{
return 0;
}
int average = sum / count;
// If we find an average return 1
if (average == K)
{
return 1;
}
}
// If we have already calculated this function
// simply return it instead of calculating it again
if (dp[index,sum,count] != -1)
{
return dp[index, sum, count];
}
// If we don't pick the current element
// simple recur for index -1
int dontpick = waysutil(index - 1,
sum, count, arr, K);
// If we pick the current element add it to
// our current sum and increment count by 1
int pick = waysutil(index - 1,
sum + arr[index],
count + 1, arr, K);
int total = pick + dontpick;
// Store the value for the current function
dp[index,sum,count] = total;
return total;
}
// Function to return the number of ways
static int ways(int N, int K, int[] arr)
{
List Arr = new List();
// Push -1 at the beginning to
// make it 1-based indexing
Arr.Add(-1);
for (int i = 0; i < N; ++i)
{
Arr.Add(arr[i]);
}
// Initialize dp array by -1
for (int i = 0; i < MAX_INDEX; i++)
{
for (int j = 0; j < MAX_SUM; j++)
{
for (int l = 0; l < MAX_INDEX; l++)
{
dp[i, j, l] = -1;
}
}
}
// Call recursive function
// waysutil to calculate total ways
int answer = waysutil(N, 0, 0, Arr, K);
return answer;
}
// Driver code
public static void Main(String []args)
{
int []arr = {3, 6, 2, 8, 7, 6, 5, 9};
int N = arr.Length;
int K = 5;
Console.WriteLine(ways(N, K, arr));
}
}
// This code is contributed by Princi Singh 输出：
19