Given two integers n and k, find how many different arrays consist of numbers from 1 to n such that there are exactly k inverse pairs.

We define an inverse pair as following: For ith and jth element in the array, if i < j and a[i] > a[j] then it's an inverse pair; Otherwise, it's not.

Since the answer may very large, the answer should be modulo 109 + 7.

Example 1:

Input: n = 3, k = 0

Output: 1

Explanation:

Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.

Example 2:

Input: n = 3, k = 1

Output: 2

Explanation:

The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Note:

1. The integer n is in the range [1, 1000] and k is in the range [0, 1000].

思路：

这种求最优解的个数，而不用返回解本身的感觉好多都用动态规划，这题也不例外。

public class Solution {

    int mo=;

    public int kInversePairs(int n, int k) {

        int[][] f=new int[][];

        f[][]=;

        for (int i=;i<=n;i++)

        {

            f[i][]=;

            for (int j=;j<=k;j++)

            {

                f[i][j]=(f[i][j-]+f[i-][j])%mo;

                if (j>=i) f[i][j]=(f[i][j]-f[i-][j-i]+mo)%mo;

            }

        }

        return f[n][k];

    }

}

参考自：

https://leetcode.com/kakahiguain/