Title: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=6012

Analysis:

Given two strings s,t 
Conclusion: the subscript of the first character with different s and t is l, and the subscript of the last character is r
         If l==r, then there is no solution
        If l < R, when we flip a substring of S, we must flip s[l] and s[r] to each other
 Counter evidence: suppose there is a reversal method of s[l] and s[r+k],
    Because s[r+k]=t[r+k]=t[l], and s[l]=t[r+k], we can get s[l]=t[l], contradiction
 l-k is the same.

So this question only needs to consider three situations:
    1. Two strings are equal: palindrome substring for s
    2.l==r, no solution
    3. For L < R, first s[l,r] turns over and is equal to t[l,r], then find the palindrome substring centered on s[l,r], otherwise there is no solution

Ac code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=5e6+5;
int p[maxn];
ll ans;
char s[maxn];
char snew[maxn];
int init(){
    int len=strlen(s),j=2;
    snew[0]='$',snew[1]='#';
    for(int i=0;i<len;i++){
        snew[j++]=s[i];
        snew[j++]='#';
    }
    snew[j]='\0';
    return 2*len+2;
}
long long manacher(){
    long long res=0;
    int len=init(),mx=0,id;
    for(int i=0;i<len;i++){
        if(mx>i)p[i]=min(p[2*id-i],mx-i);
        else p[i]=1;
        while(snew[i-p[i]]==snew[i+p[i]])///i-p[i] may be out of bounds. Use string to segment errors. char [] can
            p[i]++;
        if(mx<i+p[i])mx=i+p[i],id=i;
        res+=(long long )p[i]/2;
    }
    return res;
}
char s1[maxn],s2[maxn];
int main()
{
    //ios::sync_with_stdio(0),cin.tie(0);
    int t;
    scanf("%d%*c",&t);
    while(t--)
    {
        scanf("%s%s",s1,s2);
        int len=strlen(s1);
        int l=0x3f3f3f3f,r=0;
        for(int i=0; i<len; i++)
            if(s1[i]!=s2[i])
            {
                l=i;
                break;
            }
        for(int i=len-1;i>=0;--i)
            if(s1[i]!=s2[i])
            {
                r=i;
                break;
            }
        if(l==r)
            printf("0\n");
        else if(l>r){
            strcpy(s,s1);
            printf("%lld\n",manacher());
        }
        else if(l<r){
            ans=0;
            string sa=s1,sb=s2;
            string tmp=sa.substr(l,r-l+1);
            reverse(tmp.begin(),tmp.end());
            if(tmp==sb.substr(l,r-l+1)){
                for(int i=l-1,j=r+1;i>=0&&j<sa.size();i--,++j)
                   if(sa[i]==sa[j])
                     ++ans;
                   else break;
                printf("%lld\n",ans+1);
            }
            else puts("0");
        }
    }
    return 0;
}

 

Finally, paste a board for the longest palindrome substring from manacher:

#include <vector>
#include <iostream>
#include <string>
using namespace std;
string Manacher(string s)
{
    // Insert '#'
    string t = "$#";
    for (int i = 0; i < s.size(); ++i)
    {
        t += s[i];
        t += "#";
    }
    // Process t
    vector<int> p(t.size(), 0);///Remember to open the array more than 2 times
    int mx = 0, id = 0, resLen = 0, resCenter = 0;
    for (int i = 1; i < t.size(); ++i)  /// from 1
    {
        p[i] = mx > i ? min(p[2 * id - i], mx - i) : 1;
        while (t[i + p[i]] == t[i - p[i]]) ++p[i];
        if (mx < i + p[i])
        {
            mx = i + p[i];
            id = i;
        }
        if (resLen < p[i])  ///Find the largest p[i]
        {
            resLen = p[i];///Radius of the longest palindrome substring
            resCenter = i;///The center of the longest palindrome substring
        }
    }
    ///The longest palindrome substring length len=resLen - 1;
    ///Rescenter - reslen / 2
    return s.substr((resCenter - resLen) / 2, resLen - 1);
}
int main()
{
    int n;
    cin>>n;
    string k;
    while(n--)
    {
        cin>>k;
        cout<<Manacher(k).size()<<endl;
    }
    return 0;
}