计算总数N的可能分裂成K个整数，使得最小值至少为P

给出三个整数N，P和K，任务是找到的方式来划分N转换具有总和N，其中每个整数是≥pk信息整数总数。

例子：

Input: K = 3, N = 8, P = 2
Output: 6
Explanation:
Six possible solutions are: {2, 2, 4}, {2, 3, 3}, {2, 4, 2}, {3, 2, 3}, {3, 3, 2}, {4, 2, 2}
Each of the above integers in all combinations are ≥ P(= 2) and sum of all the combinations is N(=8).

Input: K = 2, N = 7, P = 2
Output: 4
Explanation:
Four possible solutions are: {4, 3}, {3, 4}, {5, 2}, {2, 5}
Each of the above integers in all combinations are ≥ P(= 2) and sum of all the combinations is N(= 7).

天真的方法：最简单的方法是尝试对所有具有和N的K个整数进行所有可能的组合，并检查它们是否满足上述条件。有K个位置，每个位置可以放置N个整数。因此，要检查的组合总数为N ^K。

时间复杂度： O(N ^K )
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是使用本文中讨论的方法，并且上述给定条件可以用线性方程表示：

To find the number of solutions for the equation X₁ + X₂ + X₃ + … + X_K = N where X_i ≥ P.

Assume X_i‘ = X_i – P.
Therefore, the equation reduces to:
 X₁‘ + X₂‘ + X₃‘ + … + X_K‘ = N – K*P

根据上述等式，给定的问题简化为寻找将N – K * P分解为K个整数的方法的总数。将上面找到的方法总数打印为所需答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function that finds the value of
// the Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;
 
    // Stores the value of Binomial
    // Coefficient in bottom up manner
    for (i = 0; i <= n; i++) {
 
        for (j = 0; j <= min(i, k); j++) {
 
            // Base Case
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Find the value using
            // previously stored values
            else
                C[i][j] = C[i - 1][j - 1]
                          + C[i - 1][j];
        }
    }
 
    // Return the value of C(N, K)
    return C[n][k];
}
 
// Function that count the number of
// ways to divide N into K integers
// >= P such that their sum is N
int waysToSplitN(int k, int n, int P)
{
    // Update the value of N
    int new_N = n - k * P;
 
    // Find the binomial coeffient
    // recursively
    return binomialCoeff(new_N + k - 1,
                         new_N);
}
 
// Driver Code
int main()
{
    // Given K, N, and P
    int K = 3, N = 8, P = 2;
 
    cout << waysToSplitN(K, N, P);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function that finds the value of
// the Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
  int [][]C = new int[n + 1][k + 1];
  int i, j;
 
  // Stores the value of Binomial
  // Coefficient in bottom up manner
  for (i = 0; i <= n; i++)
  {
    for (j = 0; j <= Math.min(i, k); j++)
    {
      // Base Case
      if (j == 0 || j == i)
        C[i][j] = 1;
 
      // Find the value using
      // previously stored values
      else
        C[i][j] = C[i - 1][j - 1] +
                  C[i - 1][j];
    }
  }
 
  // Return the value of C(N, K)
  return C[n][k];
}
 
// Function that count the number of
// ways to divide N into K integers
// >= P such that their sum is N
static int waysToSplitN(int k,
                        int n, int P)
{
  // Update the value of N
  int new_N = n - k * P;
 
  // Find the binomial coeffient
  // recursively
  return binomialCoeff(new_N + k - 1,
                       new_N);
}
 
// Driver Code
public static void main(String[] args)
{
  // Given K, N, and P
  int K = 3, N = 8, P = 2;
 
  System.out.print(waysToSplitN(K, N, P));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function that finds the value of
# the Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
 
    C = [[0 for x in range(k + 1)]
            for y in range(n + 1)]
 
    # Stores the value of Binomial
    # Coefficient in bottom up manner
    for i in range(n + 1):
        for j in range(min(i, k) + 1):
 
            # Base Case
            if(j == 0 or j == i):
                C[i][j] = 1
                 
            # Find the value using
            # previously stored values
            else:
                C[i][j] = (C[i - 1][j - 1] +
                           C[i - 1][j])
 
    # Return the value of C(N, K)
    return C[n][k]
 
# Function that count the number of
# ways to divide N into K integers
# >= P such that their sum is N
def waysToSplitN(k, n, P):
 
    # Update the value of N
    new_N = n - k * P
 
    # Find the binomial coeffient
    # recursively
    return binomialCoeff(new_N + k - 1,
                         new_N)
 
# Driver Code
 
# Given K, N, and P
K = 3
N = 8
P = 2
 
print(waysToSplitN(K, N, P))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
 
class GFG{
  
// Function that finds the value of
// the Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
    int [,] C = new int[n + 1, k + 1];
    int i, j;
     
    // Stores the value of Binomial
    // Coefficient in bottom up manner
    for(i = 0; i <= n; i++)
    {
        for(j = 0; j <= Math.Min(i, k); j++)
        {
             
            // Base Case
            if (j == 0 || j == i)
                C[i, j] = 1;
             
            // Find the value using
            // previously stored values
            else
                C[i, j] = C[i - 1, j - 1] +
                          C[i - 1, j];
        }
    }
     
    // Return the value of C(N, K)
    return C[n, k];
}
  
// Function that count the number of
// ways to divide N into K integers
// >= P such that their sum is N
static int waysToSplitN(int k, int n,
                        int P)
{
     
    // Update the value of N
    int new_N = n - k * P;
     
    // Find the binomial coeffient
    // recursively
    return binomialCoeff(new_N + k - 1,
                         new_N);
}
  
// Driver Code
public static void Main()
{
     
    // Given K, N, and P
    int K = 3, N = 8, P = 2;
     
    Console.Write(waysToSplitN(K, N, P));
}
}
 
// This code is contributed by sanjoy_62

输出：

6

时间复杂度： O(N ² )
辅助空间： O(N ² )