D-Double Strings
$f_{i,j}$表示a中前i个字符，b中前j个字符相同子序列的数量，容斥转移

fi,j=fi−1,j+fi,j−1−fi−1,j−1+{(1+fi−1,j−1)[ai=aj]}f_{i,j}=f_{i-1,j}+f_{i,j-1}-f_{i-1,j-1}+\{(1+f_{i-1,j-1})[a_i=a_j]\}fi,j​=fi−1,j​+fi,j−1​−fi−1,j−1​+{(1+fi−1,j−1​)[ai​=aj​]}

$g_{i,j}$表示a中前i个字符，b中前j个字符满足小于关系子序列的数量

类似相同子序列数量转移即可。
gi,j=gi−1,j+gi,j−1−gi−1,j−1+(1+gi−1,j−1)+{(fi−1,j−1+1)[ai<aj]}g_{i,j}=g_{i-1,j}+g_{i,j-1}-g_{i-1,j-1}+(1+g_{i-1,j-1})+\{(f_{i-1,j-1}+1)[a_i<a_j]\}gi,j​=gi−1,j​+gi,j−1​−gi−1,j−1​+(1+gi−1,j−1​)+{(fi−1,j−1​+1)[ai​<aj​]}

Code1

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
template <class T=int> T rd()
{T res=0;T fg=1;char ch=getchar();while(!isdigit(ch)) {if(ch=='-') fg=-1;ch=getchar();}while( isdigit(ch)) res=(res<<1)+(res<<3)+(ch^48),ch=getchar();return res*fg;
}
const int N=5010;
const ll mod=1000000007;
char a[N],b[N];
int f[N][N],g[N][N];
int n,m;
int main()
{scanf("%s%s",a+1,b+1);n=strlen(a+1);m=strlen(b+1);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(a[i]==b[j])f[i][j]=(1ll*f[i-1][j]+f[i][j-1]+1)%mod;elsef[i][j]=(1ll*mod+f[i-1][j]+f[i][j-1]-f[i-1][j-1])%mod;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(a[i]<b[j]) g[i][j]=(1ll*g[i-1][j]+g[i][j-1]+1+f[i-1][j-1])%mod;elseg[i][j]=(1ll*g[i-1][j]+g[i][j-1])%mod;}printf("%lld\n",g[n][m]);
}

Code2

首先求出相同子序列，然后枚举哪个位置不同，即第一个$a_i<a_j$，然后利用下面公式加快计算。

$$\sum _{0\le i\le k}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{k-i}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{k}\right)$$

于是有
$$\sum _{0\le i\le m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)\sum _{0\le i\le m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{m-i}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)$$

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
template <class T=int> T rd()
{T res=0;T fg=1;char ch=getchar();while(!isdigit(ch)) {if(ch=='-') fg=-1;ch=getchar();}while( isdigit(ch)) res=(res<<1)+(res<<3)+(ch^48),ch=getchar();return res*fg;
}
const int N=5010;
const int mod=1000000007;
char a[N],b[N];
int f[N][N];
int n,m;
int fac[N<<1],inv[N<<1];
ll qmi(ll a,ll b){ll v=1;while(b){if(b&1) v=v*a%mod;a=a*a%mod;b>>=1;}return v;}
void init()
{fac[0]=1;for(int i=1;i<=10000;i++) fac[i]=1ll*fac[i-1]*i%mod;inv[10000]=qmi(fac[10000],mod-2);for(int i=9999;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod;
}
int C(int n,int m){return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{init();scanf("%s%s",a+1,b+1);n=strlen(a+1);m=strlen(b+1);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(a[i]==b[j])f[i][j]=(1ll*f[i-1][j]+f[i][j-1]+1)%mod;elsef[i][j]=(1ll*mod+f[i-1][j]+f[i][j-1]-f[i-1][j-1])%mod;}ll ans=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(a[i]<b[j])ans=(ans+1ll*(f[i-1][j-1]+1)*C(n-i+m-j,n-i)%mod)%mod;printf("%lld\n",ans);
}