考虑一下插⼊法
n&lt;=100n&lt;=100n<=100
f[i][j]f[i][j]f[i][j]表⽰111~iii的全排列有j个逆序对的⽅案数
f[i][j]=Σf[i−1][j−k](0&lt;=k&lt;=i−1)f[i][j]=Σf[i-1][j-k] (0&lt;=k&lt;=i-1)f[i][j]=Σf[i1][jk](0<=k<=i1)
O(m∗n2)O(m*n^2)O(mn2)

拓展:如果n&lt;=1000n&lt;=1000n<=1000呢?

n&lt;=1000n&lt;=1000n<=1000?
f[i][j]f[i][j]f[i][j]f[i−1]f[i-1]f[i1]中连续⼀段的和
前缀和优化
O(n∗m)O(n*m)O(nm)

下面上非拓展的代码:

#include<cstdio>
using namespace std;
int f[105][6005];
int main()
{
	int n,k;
	scanf("%d%d",&n,&k);
	f[1][0]=1;
	f[2][0]=1;
	f[2][1]=1;
	f[0][0]=1;
	for(int i=3;i<=n;i++)
	{
		for(int j=0;j<=k;j++)
		{
			for(int kk=0;kk<=i-1&&kk<=j;kk++)
			{
				f[i][j]+=f[i-1][j-kk]%10000;
			}
		}
	}
	printf("%d",f[n][k]%10000);
	return 0;
}