给定两个整数N和K ，任务是计算正好具有K个反转的前N个自然数的排列数。由于计数可能非常大，请以10 9 + 7为模数打印。
An inversion is defined as a pair a[i], a[j] such that a[i] > a[j] and i < j.
例子：
Input: N = 3, K = 2
Output: 2
Explanation:
All Permutations for N = 3 are 321, 231, 213, 312, 132, 123.
Out of which only 231 and 312 have 2 inversions as:
- 231: 2 > 1 & 3 > 1
- 312: 3 > 1 & 3 > 2.
Therefore, both are satisfying the condition of having exactly K inversions.
Input: N = 5, K = 5
Output: 22
天真的方法：有关解决问题的最简单方法，请参阅上一篇文章。
时间复杂度： O(N * N！)
辅助空间： O(1)
使用自顶向下方法的动态编程：有关自顶向下方法的信息，请参阅本文的上一篇文章。
时间复杂度： O(N * K 2 )
辅助空间： O(N * K)
使用自下而上的方法进行动态编程：
插图：
For Example: N = 4, K = 2
N – 1 = 3, K0 = 0 … 123 => 1423
N – 1 = 3, K1 = 1 … 213, 132 => 2143, 1342
N – 1 = 3, K2 = 2 … 231, 312 => 2314, 3124
So the answer is 5.
The maximum value is taken between (K – N + 1) and 0 as K inversions cannot be obtained if the number of inversions in permutation of (N – 1) numbers is less than K – (N – 1) as at most (N – 1) new inversions can be obtained by adding Nth number at the beginning.
请按照以下步骤解决问题：
- 创建一个辅助数组dp [2] [K + 1] ，其中dp [N] [K]存储(N – 1)个数字的所有排列，其中K =(max(K –(N – 1)，0)到K)反演，只需将第N个数字加一次即可。
- 使用dp [i％2] [K]将在两行之间互换迭代，并取j = Max(K –(N – 1)，0) 。因此dp [N [K] = dp [N-1] [j] + dp [N-1] [j + 1] +…。 + dp [N – 1] [K] 。
- 在计算dp [N] [K]时，无需执行此额外的K迭代，因为它可以在O(1)中从dp [N] [K – 1]获得。因此，递归关系由下式给出：
dp[N][K] = dp[N][K – 1] + dp[N – 1][K] – dp[N – 1][max(K – (N – 1), 0) – 1]
- 分别在N和K上使用变量i和j迭代两个嵌套循环，并根据上述递归关系更新每个dp状态。
- 完成上述步骤后，打印dp [N％2] [K]的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count permutations with
// K inversions
int numberOfPermWithKInversion(
int N, int K)
{
// Store number of permutations
// with K inversions
int dp[2][K + 1];
int mod = 1000000007;
for (int i = 1; i <= N; i++) {
for (int j = 0; j <= K; j++) {
// If N = 1 only 1 permutation
// with no inversion
if (i == 1)
dp[i % 2][j] = (j == 0);
// For K = 0 only 1 permutation
// with no inversion
else if (j == 0)
dp[i % 2][j] = 1;
// Otherwise Update each dp
// state as per the reccurrance
// relation formed
else
dp[i % 2][j]
= (dp[i % 2][j - 1] % mod
+ (dp[1 - i % 2][j]
- ((max(j - (i - 1), 0) == 0)
? 0
: dp[1 - i % 2]
[max(j - (i - 1), 0)
- 1])
+ mod)
% mod)
% mod;
;
}
}
// Print final count
cout << dp[N % 2][K];
}
// Driver Code
int main()
{
// Given N and K
int N = 3, K = 2;
// Function Call
numberOfPermWithKInversion(N, K);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to count permutations with
// K inversions
static void numberOfPermWithKInversion(int N, int K)
{
// Store number of permutations
// with K inversions
int[][] dp = new int[2][K + 1];
int mod = 1000000007;
for(int i = 1; i <= N; i++)
{
for(int j = 0; j <= K; j++)
{
// If N = 1 only 1 permutation
// with no inversion
if (i == 1)
{
dp[i % 2][j] = (j == 0) ? 1 : 0;
}
// For K = 0 only 1 permutation
// with no inversion
else if (j == 0)
dp[i % 2][j] = 1;
// Otherwise Update each dp
// state as per the reccurrance
// relation formed
else
dp[i % 2][j] = (dp[i % 2][j - 1] % mod +
(dp[1 - i % 2][j] -
((Math.max(j - (i - 1), 0) == 0) ?
0 : dp[1 - i % 2][Math.max(j -
(i - 1), 0) - 1]) +
mod) % mod) % mod;
}
}
// Print final count
System.out.println (dp[N % 2][K]);
}
// Driver Code
public static void main(String[] args)
{
// Given N and K
int N = 3, K = 2;
// Function Call
numberOfPermWithKInversion(N, K);
}
}
// This code is contributed by akhilsainiPython3
# Python3 program for the above approach
# Function to count permutations with
# K inversions
def numberOfPermWithKInversion(N, K):
# Store number of permutations
# with K inversions
dp = [[0] * (K + 1)] * 2
mod = 1000000007
for i in range(1, N + 1):
for j in range(0, K + 1):
# If N = 1 only 1 permutation
# with no inversion
if (i == 1):
dp[i % 2][j] = 1 if (j == 0) else 0
# For K = 0 only 1 permutation
# with no inversion
elif (j == 0):
dp[i % 2][j] = 1
# Otherwise Update each dp
# state as per the reccurrance
# relation formed
else:
var = (0 if (max(j - (i - 1), 0) == 0)
else dp[1 - i % 2][max(j - (i - 1), 0) - 1])
dp[i % 2][j] = ((dp[i % 2][j - 1] % mod +
(dp[1 - i % 2][j] -
(var) + mod) % mod) % mod)
# Print final count
print(dp[N % 2][K])
# Driver Code
if __name__ == '__main__':
# Given N and K
N = 3
K = 2
# Function Call
numberOfPermWithKInversion(N, K)
# This code is contributed by akhilsainiC#
// C# program for the above approach
using System;
class GFG{
// Function to count permutations with
// K inversions
static void numberOfPermWithKInversion(int N, int K)
{
// Store number of permutations
// with K inversions
int[,] dp = new int[2, K + 1];
int mod = 1000000007;
for(int i = 1; i <= N; i++)
{
for(int j = 0; j <= K; j++)
{
// If N = 1 only 1 permutation
// with no inversion
if (i == 1)
{
dp[i % 2, j] = (j == 0) ? 1 : 0;
}
// For K = 0 only 1 permutation
// with no inversion
else if (j == 0)
dp[i % 2, j] = 1;
// Otherwise Update each dp
// state as per the reccurrance
// relation formed
else
dp[i % 2, j] = (dp[i % 2, j - 1] % mod +
(dp[1 - i % 2, j] -
((Math.Max(j - (i - 1), 0) == 0) ?
0 : dp[1 - i % 2, Math.Max(
j - (i - 1), 0) - 1]) +
mod) % mod) % mod;
}
}
// Print final count
Console.WriteLine(dp[N % 2, K]);
}
// Driver Code
public static void Main()
{
// Given N and K
int N = 3, K = 2;
// Function Call
numberOfPermWithKInversion(N, K);
}
}
// This code is contributed by akhilsaini2
时间复杂度： O(N * K)
辅助空间： O(K)