luogu2679 [NOIp2015]子串 (dp)

设f[i][j][k][b]表示在A串第i位、这是第j组、B串第k位、i号选不选（b=0/1）

那么就有$f[i][j][k][1]=(A[i]==B[k])*(f[i-1][j-1][k][0]+f[i-1][j][k-1][1]+f[i-1][j-1][k-1][1])$,$f[i][j][k][0]=f[i-1][j][k][0]+f[i-1][j][k-1][1]$

注意不要爆int，而且要开滚动数组

 #include<bits/stdc++.h>
 #define pa pair<int,int>
 #define lowb(x) ((x)&(-(x)))
 #define REP(i,n0,n) for(i=n0;i<=n;i++)
 #define PER(i,n0,n) for(i=n;i>=n0;i--)
 #define MAX(a,b) ((a>b)?a:b)
 #define MIN(a,b) ((a<b)?a:b)
 #define CLR(a,x) memset(a,x,sizeof(a))
 #define rei register int
 using namespace std;
 typedef long long ll;
 const int maxn=,maxm=,mod=1e9+;

 inline ll rd(){
     ll x=;char c=getchar();int neg=;
     while(c<''||c>''){if(c=='-') neg=-;c=getchar();}
     while(c>=''&&c<='') x=x*+c-'',c=getchar();
     return x*neg;
 }

 int N,M,K;
 int f[][maxm][maxm][],ans=;
 char A[maxn],B[maxm];

 int main(){
     // freopen("testdata.in","r",stdin);
     rei i,j,k;
     N=rd(),M=rd(),K=rd();
     scanf("%s%s",A+,B+);
     f[][][][]=(A[]==B[]);
     f[][][][]=;
     ans=f[][K][M][];
     bool b=;
     for(i=;i<=N;i++){
         for(j=;j<=min(K,i);j++){
             for(k=;k<=min(M,i);k++){
                 if(A[i]==B[k]&&j>) f[b][j][k][]=(0ll+f[b^][j-][k][]+f[b^][j][k-][]+f[b^][j-][k-][])%mod;
                 else f[b][j][k][]=;
                 f[b][j][k][]=(f[b^][j][k][]+f[b^][j][k-][])%mod;
                 // printf("%d %d %d %d %d\n",i,j,k,f[i][j][k][0],f[i][j][k][1]);
             }
         }
         if(i>=K&&i>=M) ans=(ans+f[b][K][M][])%mod;
         b^=;
     }
     printf("%d\n",ans);
     return ;
 }