I will give you two strings a and B of the same length. You can invert a substring of string a once to make a and B the same. I will ask you how many ways to invert.

It is discussed in two cases. If the initial a and B are the same, the answer is the sum of the palindrome radius of each character. If a and B are different, find out the first different position on the left and the first different position on the right respectively, and then expand the statistical answer.

AC Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
string a,b;
const int maxn=2e6+5;
int p[maxn];
void manacher()
{
    int len=a.size();
    string t="$#";
    for(int i = 0;i < len;++i)
        t+=a[i],t+='#';
    vector<int>v(t.size(),0);
    int mx=0,id=0;
    for(int i = 1;i < t.size();++i)
    {
        v[i]=mx>i?min(v[id*2-i],mx-i):1;
        while(t[i+v[i]]==t[i-v[i]])
            v[i]++;
        if(mx<i+v[i])
        mx=i+v[i],id=i;
    }
    ll ans=0;
    for(int i = 2;i < t.size();i++){
        ans+=v[i]/2;
    }
    cout<<ans<<endl;
}
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b;
        int len=a.size();
        if(a==b)
        {
            manacher();
        }
        else
        {
            int l=0,r=len-1;
            for(int i = 0;i < len;++i)
            {
                if(a[i]!=b[i])
                {
                    l=i;
                    break;
                }
            }
            for(int i = len-1;i >= 0;--i)
            {
                if(a[i]!=b[i])
                {
                    r=i;
                    break;
                }
            }
            string a1=a.substr(l,r-l+1);
            string b1=b.substr(l,r-l+1);
            reverse(a1.begin(),a1.end());
            if(a1==b1)
            {
                ll ans=1;
                while(l>0&&r<len&&(a[--l]==b[++r]))
                    ans++;
                cout<<ans<<endl;
            }
            else
            cout<<"0"<<endl;
        }
    }
    return 0;
}