NOIP2015D2T2 substring dynamic programming

Topic
This topic hits the eye
$dp\left[i\right]\left[j\right]\left[k\right]$ means the number of plans!
But think about it carefully,$The\; dp$ type does not reflect the relationship between the current character and the previous string ($dp$ formula: n You blame me for adding one dimension less?)
So we have to add another dimension$0/1$ indicates that the current character is connected to the a substring$/The$ current character is the beginning of a new substring
$0/1$ apart a little more intuitive two arrays:
We$ans\left[i\right]\left[j\right]\left[k\right]$ means to$A-$ string skewer$i$ characters$B$ string$The$ current number of substrings of$j$ characters iskkNumber of answers to$k$
$sum\left[i\right]\left[j\right]\left[k\right]$ means currently in$A-$ string skewer$i$ characters$B$ string$The$ current number of substrings of$j$ characters iskkThe number of plans at$k$ ,
then the relational expression, how to do it?
First consider the current$ans$ or use$A\left[i\right]$ or not,
so$ans\left[i\right]\left[j\right]\left[l\right]=ans[i-1]\left[j\right]\left[l\right]+sum\left[i\right]\left[j\right][l]\; is$
obvious when$A[i]\ne B[i]$时 s u m sum The value of$s$$u$$m$ is$0$
Economic intelligence$A[i]=B\left[i\right]$ At this time, either connect to the previous substring or create a new substring,
so$sum[i][j][l]=sum[i-1][j-1][l]+ans[i-1][j-1][l-1]$
Haha finished? Want to be beautifuldp [1000] [200] [200] [2]??Dp
$dthe\; p-\left[1000\right]\left[200\right]\left[200\right]\left[2\right]??$ apparently fried so we have to optimize space
noted that the first dimension looks like nothing so directly with eggs roll off the
space complexity of$O\left(4mk\right)$ Time complexity$O(nmk)\; No$ big problem
, code

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
int n,m,k,now,ans[2][202][202],sum[2][202][202];
char a[1010],b[202];
int main(){
    
    
    cin>>n>>m>>k;
    scanf("%s%s",a+1,b+1);
    ans[0][0][0]=ans[1][0][0]=now=1;
    for(int i=1;i<=n;i++){
    
    
        for(int j=1;j<=m;j++){
    
    
            for(int l=1;l<=k;l++){
    
    
                if(a[i]==b[j])sum[now][j][l]=(sum[now^1][j-1][l]+ans[now^1][j-1][l-1])%mod;
                else sum[now][j][l]=0;
                ans[now][j][l]=(ans[now^1][j][l]+sum[now][j][l])%mod;
            }
        }
        now^=1;
    }
    printf("%d\n",ans[now^1][m][k]);
}

The code is not long, huh?
$P.S.$ Since the operation caught between escape month$+The$ monthly exam is too bad$+$ Because oftoo much food, now I have no face and half full-time(It is estimated that the head teacher and coachcan notpass both levels when applying,so I can't take the exam with the second year... There is no way I can only do$NOIP\; The$ original question is to look at the knowledge points... grievances$.jpg$