我们给了N个项目，总共有K种不同的颜色。相同颜色的项目是无法区分的，并且颜色可以从1到K进行编号，每种颜色的项目计数也指定为k1，k2等。现在，对于所有可能的颜色，我们需要在颜色i的最后一项位于颜色的最后一项(i + 1)之前的约束下，逐一排列这些项。我们的目标是找出实现此目标的方法。
例子：
Input : N = 3
k1 = 1 k2 = 2
Output : 2
Explanation :
Possible ways to arrange are,
k1, k2, k2
k2, k1, k2
Input : N = 4
k1 = 2 k2 = 2
Output : 3
Explanation :
Possible ways to arrange are,
k1, k2, k1, k2
k1, k1, k2, k2
k2, k1, k1, k2
我们可以使用动态编程解决此问题。让dp [i]存储排列第i个有色项目的方式的数量。对于一种颜色的项目，答案将是一个，因为只有一种方法。现在，假设所有项目都在一个序列中。现在，要从dp [i]转到dp [i + 1]，我们需要在末尾至少放置一种颜色(i + 1)，但其他颜色(i + 1)可以放序列中的任何地方。排列颜色(i + 1)项的方式是(k1 + k2 .. + ki + k(i + 1)– 1)超过(k(i + 1)– 1)的组合，可以是表示为(k1 + k2 .. + ki + k(i + 1)– 1)C(k(i + 1)– 1)。在此表达式中我们减去了一个，因为我们需要在最后放置一个项目。
在下面的代码中，首先我们已经计算了组合值，您可以从此处了解更多信息。之后，我们遍历所有不同的颜色，并使用上述关系式计算出最终值。
C++
// C++ program to find number of ways to arrange
// items under given constraint
#include
using namespace std;
// method returns number of ways with which items
// can be arranged
int waysToArrange(int N, int K, int k[])
{
int C[N + 1][N + 1];
int i, j;
// Calculate value of Binomial Coefficient in
// bottom up manner
for (i = 0; i <= N; i++) {
for (j = 0; j <= i; 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]);
}
}
// declare dp array to store result up to ith
// colored item
int dp[K];
// variable to keep track of count of items
// considered till now
int count = 0;
dp[0] = 1;
// loop over all different colors
for (int i = 0; i < K; i++) {
// populate next value using current value
// and stated relation
dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1]);
count += k[i];
}
// return value stored at last index
return dp[K];
}
// Driver code to test above methods
int main()
{
int N = 4;
int k[] = { 2, 2 };
int K = sizeof(k) / sizeof(int);
cout << waysToArrange(N, K, k) << endl;
return 0;
} Java
// Java program to find number of ways to arrange
// items under given constraint
class GFG
{
// method returns number of ways with which items
// can be arranged
static int waysToArrange(int N, int K, int[] k)
{
int[][] C = new int[N + 1][N + 1];
int i, j;
// Calculate value of Binomial Coefficient in
// bottom up manner
for (i = 0; i <= N; i++)
{
for (j = 0; j <= i; 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]);
}
}
}
// declare dp array to store result up to ith
// colored item
int[] dp = new int[K + 1];
// variable to keep track of count of items
// considered till now
int count = 0;
dp[0] = 1;
// loop over all different colors
for (i = 0; i < K; i++)
{
// populate next value using current value
// and stated relation
dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1]);
count += k[i];
}
// return value stored at last index
return dp[K];
}
// Driver code
public static void main(String[] args)
{
int N = 4;
int[] k = new int[]{2, 2};
int K = k.length;
System.out.println(waysToArrange(N, K, k));
}
}
// This code has been contributed by 29AjayKumarPython3
# Python3 program to find number of ways
# to arrange items under given constraint
import numpy as np
# method returns number of ways with
# which items can be arranged
def waysToArrange(N, K, k) :
C = np.zeros((N + 1, N + 1))
# Calculate value of Binomial
# Coefficient in bottom up manner
for i in range(N + 1) :
for j in range(i + 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])
# declare dp array to store result
# up to ith colored item
dp = np.zeros((K + 1))
# variable to keep track of count
# of items considered till now
count = 0
dp[0] = 1
# loop over all different colors
for i in range(K) :
# populate next value using current
# value and stated relation
dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1])
count += k[i]
# return value stored at last index
return dp[K]
# Driver code
if __name__ == "__main__" :
N = 4
k = [ 2, 2 ]
K = len(k)
print(int(waysToArrange(N, K, k)))
# This code is contributed by RyugaC#
// C# program to find number of ways to arrange
// items under given constraint
using System;
class GFG
{
// method returns number of ways with which items
// can be arranged
static int waysToArrange(int N, int K, int[] k)
{
int[,] C = new int[N + 1, N + 1];
int i, j;
// Calculate value of Binomial Coefficient in
// bottom up manner
for (i = 0; i <= N; i++)
{
for (j = 0; j <= i; 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]);
}
}
// declare dp array to store result up to ith
// colored item
int[] dp = new int[K + 1];
// variable to keep track of count of items
// considered till now
int count = 0;
dp[0] = 1;
// loop over all different colors
for (i = 0; i < K; i++) {
// populate next value using current value
// and stated relation
dp[i + 1] = (dp[i] * C[count + k[i] - 1, k[i] - 1]);
count += k[i];
}
// return value stored at last index
return dp[K];
}
// Driver code
static void Main()
{
int N = 4;
int[] k = new int[]{ 2, 2 };
int K = k.Length;
Console.Write(waysToArrange(N, K, k));
}
}
// This code is contributed by DrRoot_PHP
输出：
3