Contest1818 - 2019年我能变强组队训练赛第十场 I: Isomorphic Inversion 字符串哈希

 

时间限制: 1 Sec  内存限制: 128 MB
 

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

复制样例数据

652526

样例输出

4

题解：字符串哈希，两边向中间判断即可

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6 + 10;
ull bas = 233;
ull has[N];
char s[N];
int main() {
    scanf("%s", s);
    int len = strlen(s);
    has[0] = 1;
    for(int i = 1; i <= len; i++) has[i] = has[i - 1] * bas; 
    int ans = 0;
    int i = 0, j = len - 1;
    ull cnt1, cnt2;
    int pre;
    while(i <= j) {
        cnt1 = s[i], cnt2 = s[j];
        pre = j;
    //  cout << i << " " << j << endl;
        while(cnt1 != cnt2) {
            if(i + 1 >= j - 1) break;
            i++;
            j--;
            cnt1 = cnt1 * 233 + s[i];
            cnt2 = cnt2 + s[j] * has[pre - j];
        }
        if(cnt1 != cnt2) {
            ans++;
            break;
        }
        if(i != j)
            ans += 2;
        else   
            ans++;
        i++; j--;
    }
    printf("%d\n", ans);
    return 0;   
}