Luogu P2679 substring (string +dp)

P2679 Sub-string

Test Instructions Topic Description

There are two strings that contain only lowercase letters \ (a\) and \ (b\).

Now remove \ (k\) a non-empty substring from the string \ (A\ ), and then put the \ (k\) substring in its string \ (a\) In the order in which they appear, they are concatenated together to get a new string. How many kinds of schemes can make this new string equal to the strings \ (b\) ?

Note: The location of the substring removed is also considered a different scenario.

Input output Format input format:

The first line is three positive integers \ (n,m,k\), each representing the length of the string \ (a\) , the length of the string \ (b\) , and the \ (k\)mentioned in the problem description. Each of the two integers is separated by a space.

The second line contains a string of length \ (n\) that represents the string \ (a\).

The third line contains a string of length \ (m\) that represents the string \ (b\).

Output format:

An integer that represents the number of scenarios that are being asked.

Because the answer can be large, it is required to output the result of the answer to \ (1000000007\) modulo.

Input and Output Sample input example:

6 3 1aabaabaab

Sample output:

2

Input Sample:

6 3 2aabaabaab

Sample output:

7

Input Sample:

6 3 3aabaabaab

Sample output:

7

Description

for section \ (1\) Group data:\ (1 \leq n \leq 500,1 \leq m \leq 50,k=1\);

for section \ (2\) Group to section \ (3\) Group data:\ (1 \leq n \leq 500,1 \leq m \leq 50,k=2\);

for section \ (4\) Group to section \ (5\) Group data:\ (1 \leq n \leq 500,1 \leq m \leq 50,k=m\);

for section \ (1\) Group to section \ (7\) Group data:\ (1 \leq n \leq 500,1 \leq m \leq 50,1 \leq k \leq m\);

for section \ (1\) Group to section \ (9\) Group data:\ (1 \leq n \leq 1000,1 \leq m \leq 100,1 \leq k \leq m\);

For all \ (10\) Group data:\ (1 \leq n \leq 1000,1 \leq m \leq 200,1 \leq k \leq m\).

Ideas

You can read a good blog. --alecli

This god Ben called for my problem.

Design state \ (dp[i][j][k][0/1]\),\ (i\) represents the \ (a\) string's front \ (i\) bit,\ (j\) means \ (b\) the front \ (j\) bit of the string,\ (k\) indicates how many substrings are selected,\ (0/1\) indicates that the current character is not selected in the substring.

If the bit is not selected, then the transfer is obviously easy to see:

\[dp[i][j][k][0]=dp[i-1][j-1][k][1]+dp[i-1][j-1][k][0]\]

It means that regardless of the previous selection or not, I add a space to separate the previous substring and the next substring.

If you want to choose this one, you should sort out whether the a\ and \ (b\) are the same.

• If different, then \ (dp[i][j][k][0]=0\);
• If the same, then \ (dp[i][j][k][0]=dp[i-1][j-1][k][1]+dp[i-1][j-1][k-1][1]+dp[i-1][j-1][k-1][0]\), which represents the continuation of the next substring, Restarts a new substring, directly at the beginning of a new substring, in the case of a continuous substring.

So the answer is \ (dp[n][m][k][0]+dp[n][m][k][1]\) .

By the way, my code is afraid of lack of space and wrote a scrolling array. If you do not write, remember to initialize the value of the \ (dp\) array.

AC Code

#include<bits/stdc++.h>using namespace std;typedef long long LL;const LL P=1000000007;LL n,m,k,dp[2][202][202][2];string a,b;int main(){    cin>>n>>m>>k;    cin>>a>>b;    a=' '+a;    b=' '+b;    dp[0][0][0][0]=dp[1][0][0][0]=1;    for(LL i=1;i<=n;i++)        for(LL j=1;j<=m;j++)            for(LL p=1;p<=k;p++)            {                dp[i&1][j][p][0]=(dp[(i-1)&1][j][p][0]+dp[(i-1)&1][j][p][1])%P;                if(a[i]==b[j]) dp[i&1][j][p][1]=(dp[(i-1)&1][j-1][p][1]+dp[(i-1)&1][j-1][p-1][0]+dp[(i-1)&1][j-1][p-1][1])%P;                else dp[i&1][j][p][1]=0;            }    printf("%lld",(dp[n&1][m][k][1]+dp[n&1][m][k][0])%P);    return 0;}

Luogu P2679 substring (string +dp)