LeetCode 629. K个逆序对数组

 1 class Solution {
 2     int dp[2][1007], mod = 1e9 + 7, now = 0;
 3 public:
 4     int kInversePairs(int n, int k) {
 5         dp[now][0] = 1;
 6         
 7         for(int i = 1; i <= n; ++i) {
 8             now = !now;
 9             dp[now][0] = 1;
10             for(int j = 1; j <= k; ++j) {
11                 if(2 * j > i * (i - 1)) break;
12                 dp[now][j] = (dp[now][j - 1] + dp[!now][j]) % mod;
13                 if(j - i >= 0) dp[now][j] = (dp[now][j] - dp[!now][j - i] + mod) % mod;
14             }
15         }
16         
17         return dp[now][k];
18     }
19 };