给你两个子串，让你找出来一个最短的字符串包括这两个子串，输出最多的子串有多少种。

类似于最长公共子序列，相等的话长度+1，不想等的话比較长度，使用长度小的。

1577. E-mail

b

ab

1

abcab

cba

4

#include <algorithm>

#include <iostream>

#include <stdlib.h>

#include <string.h>

#include <iomanip>

#include <stdio.h>

#include <string>

#include <queue>

#include <cmath>

#include <stack>

#include <map>

#include <set>

#define eps 1e-8

#define M 1000100

#define LL long long

//#define LL long long

#define INF 0x3f3f3f

#define PI 3.1415926535898

#define mod 1000000007

const int maxn = 2010;

using namespace std;

LL dp[maxn][maxn];

LL len[maxn][maxn];

char str1[maxn];

char str2[maxn];

int main()

{

    while(~scanf("%s %s",str1+1, str2+1))

    {

        int n = strlen(str1+1);

        int m = strlen(str2+1);

        memset(dp, 0, sizeof(dp));

        memset(len, 0, sizeof(len));

        for(int i = 0; i <= max(n, m); i++)

        {

            dp[i][0] = 1;

            dp[0][i] = 1;

            len[i][0] = i;

            len[0][i] = i;

        }

        for(int i = 1; i <= n; i++)

        {

            for(int j = 1; j <= m; j++)

            {

                if(str1[i] == str2[j])

                {

                    dp[i][j] = dp[i-1][j-1];

                    len[i][j] = len[i-1][j-1]+1;

                    continue;

                }

                if(len[i-1][j] > len[i][j-1])

                {

                    dp[i][j] = dp[i][j-1];

                    len[i][j] = len[i][j-1]+1;

                    continue;

                }

                if(len[i][j-1] > len[i-1][j])

                {

                    dp[i][j] = dp[i-1][j];

                    len[i][j] = len[i-1][j]+1;

                    continue;

                }

                len[i][j] = len[i-1][j]+1;

                dp[i][j] += dp[i-1][j]+dp[i][j-1];

                dp[i][j] %= mod;

            }

        }

        cout<<dp[n][m]<<endl;

    }

    return 0;

}