给定大小为N且整数K的数组arr [] ，任务是找到此数组的K个长度子序列的数量，以使这些子序列的总和最小。
例子：
Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 1
Subsequences of lengths 2 are (1, 2), (1, 3), (1, 4),
(2, 3), (2, 4) and (3, 4).
The minimum sum is 3 and the only subsequence
with this sum is (1, 2).
Input: arr[] = {2, 1, 2, 2, 2, 1}, K = 3
Output: 4
方法：给定数组中长度为K的子序列的最小可能总和是数组中K个最小元素的总和。令X为数组的K个最小元素中的最大元素，令它在数组的K个最小元素中出现的次数为Y ，其在整个数组中的总出现次数为cntX 。现在，有cntX C Y种方法可以在K个最小元素中选择该元素，这是所需子序列的数量。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the value of
// Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
int C[n + 1][k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= min(i, k); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
return C[n][k];
}
// Function to return the count
// of valid subsequences
int cntSubSeq(int arr[], int n, int k)
{
// Sort the array
sort(arr, arr + n);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--) {
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++) {
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof(int);
int k = 2;
cout << cntSubSeq(arr, n, k);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
int C[][] = new int [n + 1][k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.min(i, k); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] +
C[i - 1][j];
}
}
return C[n][k];
}
// Function to return the count
// of valid subsequences
static int cntSubSeq(int arr[], int n, int k)
{
// Sort the array
Arrays.sort(arr);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--)
{
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++)
{
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int n = arr.length;
int k = 2;
System.out.println(cntSubSeq(arr, n, k));
}
}
// This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
int [,]C = new int [n + 1, k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, k); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i, j] = 1;
// Calculate value using previously
// stored values
else
C[i, j] = C[i - 1, j - 1] +
C[i - 1, j];
}
}
return C[n, k];
}
// Function to return the count
// of valid subsequences
static int cntSubSeq(int []arr, int n, int k)
{
// Sort the array
Array.Sort(arr);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--)
{
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++)
{
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 1, 2, 3, 4 };
int n = arr.Length;
int k = 2;
Console.WriteLine(cntSubSeq(arr, n, k));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to return the value of
# Binomial Coefficient C(n, k)
def binomialCoeff(n, k) :
C = [[0 for i in range(n + 1)]
for j in range(k + 1)]
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range (0, n + 1 ):
for j in range (0, min(i, k) + 1):
# Base Cases
if (j == 0 or j == i):
C[i][j] = 1
# Calculate value using previously
# stored values
else :
C[i][j] = C[i - 1][j - 1] + C[i - 1][j]
return C[n][k]
# Function to return the count
# of valid subsequences
def cntSubSeq(arr, n, k) :
# Sort the array
arr.sort()
# Maximum among the minimum K elements
num = arr[k - 1];
# Y will store the frequency of num
# in the minimum K elements
Y = 0;
for i in range (k - 1, -1, 1) :
if (arr[i] == num):
Y += 1
# cntX will store the frequency of
# num in the complete array
cntX = Y;
for i in range (k, n):
if (arr[i] == num) :
cntX += 1
return binomialCoeff(cntX, Y)
# Driver code
arr = [ 1, 2, 3, 4 ]
n = len(arr)
k = 2
print(cntSubSeq(arr, n, k))
# This code is contributed by ihritik输出：
1