问题 I: Isomorphic Inversion

时间限制: 1 Sec 内存限制: 128 MB

提交 状态 题面

题目描述

Let s be a given string of up to 106 digits. Find the maximal k for which it is possible to partition s into k consecutive contiguous substrings, such that the k parts form a palindrome.

More precisely, we say that strings s0, s1, . . . , sk−1 form a palindrome if si = sk−1−i for all 0 ≤ i < k.

In the first sample case, we can split the string 652526 into 4 parts as 6|52|52|6, and these parts together form a palindrome. It turns out that it is impossible to split this input into more than 4 parts while still making sure the parts form a palindrome.

输入

A nonempty string of up to 106 digits.

输出

Print the maximal value of k on a single line.

样例输入

652526

样例输出

4

字符串哈希裸题，O(n)

#pragma GCC optimize(3,"Ofast","inline")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
#include <map>
#include <cmath>
#include <stack>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1010;
const int MAXN = 1000010;
const int Hash = 13131;
const int mod =1e9+7;
ull base[MAXN],h[MAXN];
char str[MAXN];
void init(int n){
    base[0]=1;
    h[0]=Hash;
    for(int i=1;i<=n;i++){
        base[i]=base[i-1]*Hash;
        h[i]=h[i-1]*Hash+str[i];
    }
}
bool check(int l,int r,int len){
    //cout<<"checkl "<<l<<" "<<len<<" "<<h[l]-h[l-len]*base[len]<<endl;
    //cout<<"checkr "<<r<<" "<<len<<" "<<h[r+len-1]-h[r-1]*base[len]<<endl;
    return h[l]-h[l-len]*base[len]==h[r+len-1]-h[r-1]*base[len];
}
int main()
{
    scanf("%s",str+1);
    int len=strlen(str+1);
   // cout<<len<<endl;
    init(len);
    int ans=0;
    int anslen=0;
    for(int i=1,j=len;i<=(len>>1)&&j>=(len>>1);i++,j--){
        //cout<<i<<" "<<anslen<<endl;
        anslen++;
        if(check(i,j,anslen)){
            ans++;
            anslen=0;
        }
    }
    ans<<=1;
    //if(anslen==0)
    //
    //    ans+=len&1;
    if(len&1){
        ans++;
    }else{
        if(anslen!=0)
            ans++;
    }
    cout<<ans<<endl;
    return 0;
}