给定两个正整数N和K。任务是找到可以形成的大小为N的数组的数量，以使数组的元素应为正整数，并且元素的总和等于K。
例子：
Input : N = 2, K = 3
Output : 2
Explanation: [1, 2] and [2, 1] are the only arrays of size 2 whose sum is 3.
Input : n = 3, k = 7
Output : 15
先决条件：星条旗
假设有K个相同的对象需要放置在N个容器(数组的N个索引)中，以使每个容器中至少有一个对象。我们开始将对象放在一行上，而不是开始将对象放到垃圾箱中，第一个垃圾箱的对象将从左侧开始，然后是第二个垃圾箱的对象，依此类推。因此，一旦知道第一个对象进入第二个容器，第一个对象进入第三个容器，等等，就将确定配置。我们可以通过在两个对象之间的某些位置放置NX 1分隔条来表明这一点。由于不允许将垃圾箱为空，因此在给定的一对对象之间最多可以有一个条形。因此，我们有K个对象与K – 1个间隙成一条直线。现在我们必须选择N – 1个间隙以放置K – 1个间隙中的条。可以通过K – 1 C N – 1来选择。
以下是此方法的实现：
C++
// CPP Program to find the number of arrays of
// size N whose elements are positive integers
// and sum is K
#include
using namespace std;
// Return nCr
int binomialCoeff(int n, int k)
{
int C[k + 1];
memset(C, 0, sizeof(C));
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++) {
// Compute next row of pascal triangle using
// the previous row
for (int j = min(i, k); j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// Return the number of array that can be
// formed of size n and sum equals to k.
int countArray(int N, int K)
{
return binomialCoeff(K - 1, N - 1);
}
// Driver Code
int main()
{
int N = 2, K = 3;
cout << countArray(N, K) << endl;
return 0;
} Java
// Java Program to find the
// number of arrays of size
// N whose elements are positive
// integers and sum is K
import java.io.*;
class GFG
{
// Return nCr
static int binomialCoeff(int n,
int k)
{
int []C = new int[k + 1];
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++)
{
// Compute next row of pascal
// triangle using the previous row
for (int j = Math.min(i, k); j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// Return the number of
// array that can be
// formed of size n and
// sum equals to k.
static int countArray(int N, int K)
{
return binomialCoeff(K - 1, N - 1);
}
// Driver Code
public static void main (String[] args)
{
int N = 2, K = 3;
System.out.println( countArray(N, K));
}
}
// This code is contributed by anuj_67.Python3
# Python3 Program to find the number
# of arrays of size N whose elements
# are positive integers and sum is K
# Return nCr
def binomialCoeff(n, k):
C = [0] * (k + 1);
C[0] = 1; # nC0 is 1
for i in range(1, n + 1):
# Compute next row of pascal
# triangle using the previous row
for j in range(min(i, k), 0, -1):
C[j] = C[j] + C[j - 1];
return C[k];
# Return the number of array that
# can be formed of size n and
# sum equals to k.
def countArray(N, K):
return binomialCoeff(K - 1, N - 1);
# Driver Code
N = 2;
K = 3;
print(countArray(N, K));
# This code is contributed by mitsC#
// C# Program to find the
// number of arrays of size
// N whose elements are positive
// integers and sum is K
using System;
class GFG
{
// Return nCr
static int binomialCoeff(int n,
int k)
{
int []C = new int[k + 1];
C[0] = 1; // nC0 is 1
for (int i = 1; i <= n; i++)
{
// Compute next row of
// pascal triangle using
// the previous row
for (int j = Math.Min(i, k);
j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
// Return the number of
// array that can be
// formed of size n and
// sum equals to k.
static int countArray(int N,
int K)
{
return binomialCoeff(K - 1,
N - 1);
}
// Driver Code
static public void Main ()
{
int N = 2, K = 3;
Console.WriteLine(
countArray(N, K));
}
}
// This code is contributed by ajitPHP
0; $j--)
$C[$j] = $C[$j] +
$C[$j - 1];
}
return $C[$k];
}
// Return the number of
// array that can be
// formed of size n and
// sum equals to k.
function countArray($N, $K)
{
return binomialCoeff($K - 1,
$N - 1);
}
// Driver Code
$N = 2;
$K = 3;
echo countArray($N, $K);
// This code is contributed by mits
?>输出：
2