【noi 2.6_9289】&【bzoj2023 / 1630】Ant Counting 数蚂蚁｛Usaco2005 Nov｝

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 using namespace std;
 7 #define M 1010
 8 #define N 100010
 9 #define mod 1000000
10 
11 int a[N],h[M],f[2][N];//h[i]就是题目中的Ni
12 int mmin(int x,int y) {return x<y?x:y;}
13 
14 int main()
15 {
16     int m,n,l,r;
17     scanf("%d%d%d%d",&m,&n,&l,&r);
18     for (int i=1;i<=n;i++) scanf("%d",&a[i]);
19     sort(a+1,a+1+n);
20     int t=1; h[1]=1,h[m+1]=n+1;
21     for (int i=2;i<=n;i++)
22       if (a[i]!=a[i-1]) h[++t]=i;
23   
24     f[0][0]=1;
25     for (int j=1;j<h[2];j++) f[0][j]=0;
26     int u=1;
27     for (int i=1;i<=m;i++)
28     {
29       f[u][0]=1;
30       for (int j=1;j<h[i+1];j++)
31       {
32         int k=mmin(h[i+1]-h[i],j);
33         f[u][j]=((f[u][j-1]+f[1-u][j])%mod-f[1-u][j-k-1]%mod+mod)%mod;
34         //for (int k=0;k<=h[i+1]-h[i]&&k<=j;k++)
35         //  f[i][j]+=f[i-1][j-k];
36       }
37       u=1-u;
38     }
39     int ans=0;
40     for (int j=l;j<=r;j++) ans=(ans+f[1-u][j])%mod;
41     printf("%d\n",ans);
42     return 0;
43 }