SPOJ-EPALIN 字符串哈希 回文

题意：给一个串s，用最少的字母补全使它成为回文串

思路：先求正反两个哈希，然后枚举终点

思维上没什么难度主要是细节以及怎么写比较自然

我选择把两个哈希串都弄成递增的，那原串s[l..r]的哈希值就是反哈希[n-r, n-l]的值

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define uLL unsigned long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

const int mx = 1e5+10;
const int base = 131;
uLL hs[mx], ht[mx], p[mx];
char s[mx];

int main(){
    p[0] = 1;
    for (int i = 1; i < mx; i++) p[i] = p[i-1]*base;
    while (scanf("%s", s+1) == 1){
        int n = strlen(s+1), flag = -1;
        for (int i = 1; i <= n; i++) {
            hs[i] = hs[i-1]*base+s[i]-'a';
            ht[i] = ht[i-1]*base+s[n-i+1]-'a';
        }
        int i = n/2+1;
        while (i <= n){
            uLL a = hs[n]-p[n-i+1]*hs[i-1];
            uLL b = ht[2*n-2*i+2]-p[n-i+1]*ht[n-i+1];
            if (2*i-n >= 0 && a == b){
                flag = 0;
                break;
            }
            a = hs[n]-p[n-i]*hs[i];
            b = ht[2*n-2*i+1]-p[n-i]*ht[n-i+1];
            if (i+1 <= n && a == b) {
                flag = 1;
                break;
            }
            i++;
        }
        printf("%s", s+1);
        if (flag == 1){
            for (int j = 2*i-n-1; j >= 1; j--)
                printf("%c", s[j]);
        }
        else if (!flag){
            for (int j = 2*i-n-2; j >= 1; j--)
                printf("%c", s[j]);
        }
        else {
            for (int j = n-1; j >= 1; j--)
                printf("%c", s[j]);
        }
        printf("
");
    }
    return 0;
}