题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2431 很容易想到n^3的做法.就是前 i 个数用第 i 个数最多能 i - 1 个逆序对,所以 i - 1 个数中属于 j ~ j - i + 1 的值都能加到前 i 个数的状态上. #include<iostream> #include<cstdio> #include<cstring> ,mod=; int n,k,dp[N][N],sm[N]; int ma…
f[i][j]前i个数有j个逆序对的数量 f[i][j]=sigma(f[i-1][j-k]){1<=k<=i} 维护一个前缀和即可 #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<algorithm> #define ll long long using namespace std; ,mod=1e4; int n,k;…