"Noi 2.6_9289" & "bzoj2023/1630" ant counting number ant {Usaco2005 Nov}

Test instructions: An ant with a family of M, each NI only (the same as each other). Ask how many different combinations of l~r are selected.

Solution: A very basic DP, do not be "permutation combination" confused ah ~ I have been exposed to this type, but also forgot, must remember!
This is a type of DP--M species N for DP, which defines F[I][J] to denote the number of schemes of J ("Only" ants) selected in the first I species (this is "family"). Then the layered DP.

So F[i][j]=sum{f[i-1][j-k]} (0<=K<=NI), then use prefix and optimize time + scroll array to optimize the space.

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7 #defineM 10108 #defineN 1000109 #defineMoD 1000000Ten  One inta[n],h[m],f[2][n];//H[i] is the problem of NI A intMmin (intXintY) {returnX<y?x:y;} -  - intMain () the { -     intM,n,l,r; -scanf"%d%d%d%d",&m,&n,&l,&R); -      for(intI=1; i<=n;i++) scanf ("%d",&a[i]); +Sort (A +1, A +1+n); -     intt=1; h[1]=1, h[m+1]=n+1; +      for(intI=2; i<=n;i++) A       if(a[i]!=a[i-1]) h[++t]=i; at    -f[0][0]=1; -      for(intj=1; j2];j++) f[0][j]=0; -     intu=1; -      for(intI=1; i<=m;i++) -     { inf[u][0]=1; -        for(intj=1; j1];j++) to       { +         intK=mmin (h[i+1]-h[i],j); -F[u][j]= ((f[u][j-1]+f[1-U][J])%mod-f[1-u][j-k-1]%MOD+MOD)%MoD; the         //for (int k=0;k<=h[i+1]-h[i]&&k<=j;k++) *         //F[i][j]+=f[i-1][j-k]; $       }Panax Notoginsengu=1-u; -     } the     intans=0; +      for(intj=l;j<=r;j++) ans= (ans+f[1-U][J])%MoD; Aprintf"%d\n", ans); the     return 0; +}

"Noi 2.6_9289" & "bzoj2023/1630" ant counting number ant {Usaco2005 Nov}