字符串哈希

1.区域赛银牌:Problem - J - Codeforces

这道题首先覆盖了字符串哈希的基础板子，其次，通过pll add 和mul函数来重新计算区间反转后的hash值，然后再判断回文

字符串哈希的实质的递推求每一项，求区间子串哈希利用前缀和思想

#include <bits/stdc++.h>  //�ص������㰮�ҵĶ���
using namespace std;
#define ll long long
#define pll pair<long, long>
const int MAXN = 1e6 + 5;

struct StringHash {
  int n;           // 字符串长度
  char str[MAXN];  // 下标从1开始
  const ll Base1 = 29, MOD1 = 1e9 + 7;
  const ll Base2 = 131, MOD2 = 1e9 + 9;
  ll ha1[MAXN], ha2[MAXN];    // 正着的哈希值
  ll rha1[MAXN], rha2[MAXN];  // 反着的哈希值
  ll pow1[MAXN], pow2[MAXN];  // Base1和Base2的乘方

  void init() {  // 预处理pow1[]、pow2[]
    pow1[0] = pow2[0] = 1;
    for (int i = 1; i <= n; i++) {
      pow1[i] = pow1[i - 1] * Base1 % MOD1;
      pow2[i] = pow2[i - 1] * Base2 % MOD2;
    }
  }

  void pre() {  // 预处理ha1[]、ha2[]
    for (int i = 1; i <= n; i++) {
      ha1[i] = (ha1[i - 1] * Base1 + str[i]) % MOD1;
      ha2[i] = (ha2[i - 1] * Base2 + str[i]) % MOD2;
      rha1[i] = (rha1[i - 1] * Base1 + str[n - i + 1]) % MOD1;
      rha2[i] = (rha2[i - 1] * Base2 + str[n - i + 1]) % MOD2;
    }
  }

  pll get_hash(int l, int r) {  // 求子串str[l...r]正着的哈希值
    ll res1 = ((ha1[r] - ha1[l - 1] * pow1[r - l + 1]) % MOD1 + MOD1) % MOD1;
    ll res2 = ((ha2[r] - ha2[l - 1] * pow2[r - l + 1]) % MOD2 + MOD2) % MOD2;
    return pll(res1, res2);
  }

  pll get_rhash(int l, int r) {  // 求子串str[l...r]反着的哈希值
    ll res1 =
        ((rha1[n - l + 1] - rha1[n - r] * pow1[r - l + 1]) % MOD1 + MOD1) %
        MOD1;
    ll res2 =
        ((rha2[n - l + 1] - rha2[n - r] * pow2[r - l + 1]) % MOD2 + MOD2) %
        MOD2;
    return pll(res1, res2);
  }

  bool IsPalindrome(int l, int r) {  // 判断子串str[l...r]是否是回文串
    return get_hash(l, r) == get_rhash(l, r);
  }

  pll add(pll a, pll b) {
    ll res1 = (a.first + b.first) % MOD1;
    ll res2 = (a.second + b.second) % MOD2;
    return pll(res1, res2);
  }

  pll mul(pll& a, ll k) {  // a *= Base的k次方
    ll res1 = a.first * pow1[k] % MOD1;
    ll res2 = a.second * pow2[k] % MOD2;
    return pll(res1, res2);
  }
} solver;

void solve() {
  cin >> solver.str + 1;

  solver.n = strlen(solver.str + 1);
  solver.init();
  solver.pre();

  int l = 1, r = solver.n;
  while (l < r && solver.str[l] == solver.str[r])
    l++, r--;  // 去掉两边相同的字符

  if (l >= r) {  // 原串是回文串
    cout << "1 1" << endl;
    return;
  }

  for (int i = l; i <= r; i++) {  // 翻转前缀str[l...i]
    auto tmp1 = solver.get_rhash(l, i), tmp2 = solver.get_hash(i + 1, r);
    auto res1 = solver.add(solver.mul(tmp1, r - i), tmp2);
    auto tmp3 = solver.get_hash(l, i), tmp4 = solver.get_rhash(i + 1, r);
    auto res2 = solver.add(solver.mul(tmp4, i - l + 1), tmp3);
    if (res1 == res2) {
      cout << l << ' ' << i;
      return;
    }
  }

  for (int i = l; i <= r; i++) {  // 反转后缀str[i...r]
    auto tmp1 = solver.get_hash(l, i - 1), tmp2 = solver.get_rhash(i, r);
    auto res1 = solver.add(solver.mul(tmp1, r - i + 1), tmp2);
    auto tmp3 = solver.get_rhash(l, i - 1), tmp4 = solver.get_hash(i, r);
    auto res2 = solver.add(solver.mul(tmp4, i - l), tmp3);
    if (res1 == res2) {
      cout << i << ' ' << r;
      return;
    }
  }

  cout << "-1 -1";
}

int main() { solve(); }

 

 

练习: (稍后更新)