Codeforces Global Round 7 D2. Prefix-Suffix Palindrome (Hard version)(hash)

D2. Prefix-Suffix Palindrome (Hard version)
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.

You are given a string s, consisting of lowercase English letters. Find the longest string, t, which satisfies the following conditions:

The length of t does not exceed the length of s.
t is a palindrome.
There exists two strings a and b (possibly empty), such that t=a+b ( “+” represents concatenation), and a is prefix of s while b is suffix of s.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤105), the number of test cases. The next t lines each describe a test case.

Each test case is a non-empty string s, consisting of lowercase English letters.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 106.

Output
For each test case, print the longest string which satisfies the conditions described above. If there exists multiple possible solutions, print any of them.

Example
inputCopy
5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba
outputCopy
a
abcdfdcba
xyzyx
c
abba
Note
In the first test, the string s=“a” satisfies all conditions.

In the second test, the string “abcdfdcba” satisfies all conditions, because:

Its length is 9, which does not exceed the length of the string s, which equals 11.
It is a palindrome.
“abcdfdcba” = “abcdfdc” + “ba”, and “abcdfdc” is a prefix of s while “ba” is a suffix of s.
It can be proven that there does not exist a longer string which satisfies the conditions.

In the fourth test, the string “c” is correct, because “c” = “c” + “” and a or b can be empty. The other possible solution for this test is “s”.

题意 :
给一个字符串，找到字符串的一个前缀和后缀（可以为空） 使这两个串组成一个回文串。

思路 ：
先让两端的字符串进行匹配 直到两端的字符不同 。
然后从前后两个位置开始找到最长的两个回文串 s1 s2
最后得到的答案就是 已经匹配的部分 中间插入 max(s1,s2);
找回文的过程用 hash 优化

#include<iostream>
#include<cstdio>
#include<map>
#include<math.h>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
const int rr=2333;
const int mod=999999797;
ll pre[maxn],suf[maxn],poww[maxn];
char str[maxn];
string s,s1,s2,ans;
int main (){
    ll k,len,t,hashp,hashs;;
    cin>>k;
    while(k--){
       ll l=0,r=0,lll=0;
       scanf("%s",str+1);
       len=strlen(str+1);
       pre[0]=0;suf[len+1]=0,t=0,s="",poww[0]=1,ans="",s1="",s2="";
       for(int i=1;i<=len;i++) poww[i]=poww[i-1]*rr%mod;                //记录 rr的 n次方
       for(int i=1;i<=len;i++) pre[i]=(pre[i-1]*rr+(str[i]-'a'+1))%mod; //前缀 hash
       for(int i=len;i>=1;i--) suf[i]=(suf[i+1]*rr+(str[i]-'a'+1))%mod; //后缀 hash
       
       while(str[1+t]==str[len-t]&&t<len/2-1){          //找到第一个不匹配的位置
            s+=str[1+t];
            t++;
       }
       
       l=t+1,r=t+1;
       for(int i=1+t;i<=len-t;i++){
            hashp=(pre[i]-pre[l-1]*poww[i-l+1]%mod+mod)%mod;
            hashs=(suf[l]-suf[i+1]*poww[i-l+1]%mod+mod)%mod;
           if(hashp==hashs&&i-l+1>lll){       //若是回文 （前后 hash 相同） 记录 回文串的两个端点
              l=t+1;r=i;lll=r-l+1;
           }
       }
       for(int i=l;i<=r;i++) s1+=str[i];      //记录 从左至右的最大回文
       
       r=len-t,l=len-t;
       for(int i=len-t;i>=1+t;i--){
            hashp=(pre[r]-pre[i-1]*poww[r-i+1]%mod+mod)%mod;
            hashs=(suf[i]-suf[r+1]*poww[r-i+1]%mod+mod)%mod;
           if(hashp==hashs&&r-i+1>lll){
              l=i;r=len-t;lll=r-l+1;
           }
       }
       for(int i=l;i<=r;i++) s2+=str[i];      //记录 从右至左的最大回文

       if(s1.size()>s2.size()) ans=s+s1,reverse(s.begin(),s.end()),ans+=s;
       else ans=s+s2,reverse(s.begin(),s.end()),ans+=s;

       cout<<ans<<endl;





    }






}
//999999797      999999893  999998639;