leetcode：K inverse pairs array

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 = 0Output: 1Explanation: 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 = 1Output: 2Explanation: 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].

给了1到n总共n个数字，任意排列数组的顺序，使其刚好存在k个翻转对，

而且题目中表结果会很大，要对一个很大的数字取余。对于结果巨大的题目,最先应该考虑的就是DP。

需要一个二维的DP数组，其中dp[i][j]表示1到i的数字中有j个翻转对的排列总数，要求的就是dp[n][k]了，即1到n的数字中有k个翻转对的排列总数。

dp[n+1][k]是1到n+1点数字中有k个翻转对的个数，那么实际上在1到n的数字中的某个位置加上了n+1这个数

dp[n][k] = dp[n - 1][k] + dp[n - 1][k-1] + ... + dp[n - 1][k - n + 1]

class Solution {public:    int kInversePairs(int n, int k) {        int M = 1000000007;        vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));        dp[0][0] = 1;        for (int i = 0; i <= n; ++i) {            for (int j = 0; j < i; ++j) {                for (int m = 0; m <= k; ++m) {                    if (m - j >= 0 && m - j <= k) {                        dp[i][m] = (dp[i][m] + dp[i - 1][m - j]) % M;                    }                }            }        }        return dp[n][k];    }};