bzoj 2431: [HAOI2009]逆序对数列

一个比较弱智的DP

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<vector>
 9 #define M 1009
10 #define EPS 1e-10
11 #define MO 10000
12 #define ll long long
13 using namespace std;
14 ll read()
15 {
16     char ch=getchar();
17     ll x=0,f=1;
18     for(;ch<‘0‘||ch>‘9‘;ch=getchar())
19         if(ch==‘-‘)
20           f=-1;
21     for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())
22         x=x*10+ch-‘0‘;
23     return x*f;
24 }
25 int n,k,f[M][M],sum[M][M];
26 int main()
27 {
28    n=read();
29    k=read();
30    for(int i=0;i<=n;i++)
31      f[i][0]=sum[i][0]=1;
32    for(int i=1;i<=n;i++)
33      for(int j=1;j<=min(k,(i-1)*i/2);j++)
34        {
35          if(j-i>=0)
36            f[i][j]=(sum[i-1][min(j,(i-2)*(i-1)/2)]-sum[i-1][max(0,j-i)]+MO)%MO;
37          else
38            f[i][j]=sum[i-1][min(j,(i-2)*(i-1)/2)]%MO;
39          sum[i][j]=(sum[i][j-1]+f[i][j])%MO;
40        }
41     printf("%d\n",f[n][k]);
42    return 0;
43 }
44