例题：https://acm.sdut.edu.cn/onlinejudge3/problems/4928ui​​​​​​

第一种方式：

这一种做法就和正常的字符串哈希基本思想一样，都是h[r]=h[l-1]*p[r-l+1];

我们这里是左高右低，故是hash[ l ]=hash[ l ]*p[ r.len ]+hash[ r ];

且注意要query要先递归右子树，以便于先维护出右部分的len

#include <bits/stdc++.h>
using namespace std;
#define pi acos(-1)
#define xx first
#define yy second
#define endl "\n"
#define lowbit(x) x & (-x)
#define int long long
#define ull unsigned long long
#define pb push_back
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define Ysanqian ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
const int N = 1e6 + 10, M = 1010, inf = 0x3f3f3f3f, mod = 998244353, P = 13331;
const double eps = 1e-8;
char s[N];
int n, q;
int p[N];
struct node
{int l, r;int len;int hash1, hash2;
} tr[N << 2];
void init()
{p[0] = 1;for (int i = 1; i <= n + 10; i++)p[i] = p[i - 1] * P;
}
void pushup(node &u, node &l, node &r)
{u.hash1 = l.hash1 * p[r.len] + r.hash1;u.hash2 = r.hash2 * p[l.len] + l.hash2;
}
void pushup(int u)
{pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void build(int u, int l, int r)
{tr[u] = {l, r, r - l + 1, 0, 0};if (l == r){tr[u] = {l, r, 1, (int)s[l], (int)s[l]};return;}int mid = l + r >> 1;build(u << 1, l, mid);build(u << 1 | 1, mid + 1, r);pushup(u);
}
void modify(int u, int x, int k)
{if (tr[u].l == x && tr[u].r == x){tr[u].hash1 = k;tr[u].hash2 = k;return;}int mid = tr[u].l + tr[u].r >> 1;if (x <= mid)modify(u << 1, x, k);elsemodify(u << 1 | 1, x, k);pushup(u);
}
int query1(int u, int l, int r, int &len1)
{if (tr[u].l >= l && tr[u].r <= r){int temp = (tr[u].hash1 * p[len1]) % mod;len1 += tr[u].len;return temp;}else{int res = 0;int mid = tr[u].l + tr[u].r >> 1;if (r > mid)res += query1(u << 1 | 1, l, r, len1);if (l <= mid)res += query1(u << 1, l, r, len1);return res % mod;}
}
int query2(int u, int l, int r, int &len2)
{if (tr[u].l >= l && tr[u].r <= r){int temp = (tr[u].hash2 * p[len2]) % mod;len2 += tr[u].len;return temp;}int res = 0;int mid = tr[u].l + tr[u].r >> 1;if (l <= mid)res += query2(u << 1, l, r, len2);if (r > mid)res += query2(u << 1 | 1, l, r, len2);return res % mod;
}
void solve()
{cin >> n >> q;cin >> (s + 1);init();build(1, 1, n);while (q--){int op;cin >> op;if (op == 2){int l, r;cin >> l >> r;int len1 = 0, len2 = 0;int x = query1(1, l, r, len1);int y = query2(1, l, r, len2);if (x == y)cout << "Yes" << endl;elsecout << "No" << endl;}else{int x;char k;cin >> x >> k;modify(1, x, (int)k);}}
}
signed main()
{Ysanqian;int T;T = 1;// cin >> T;while (T--)solve();return 0;
}

第二种方式：

这一种直接暴力维护的正反哈希，一个左高右低，一个右高左低，相比较于第一种做法多了一个快速幂的log( 注意取mod，不然会wa）

#include <bits/stdc++.h>
using namespace std;
#define pi acos(-1)
#define xx first
#define yy second
#define endl "\n"
#define lowbit(x) x & (-x)
#define int long long
#define ull unsigned long long
#define pb push_back
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define Ysanqian ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
const int N = 1e6 + 10, M = 1010, inf = 0x3f3f3f3f, mod = 1e9 + 7, P = 13331;
const double eps = 1e-8;
char s[N];
int n, q;
int p[N];
struct node
{int l, r;int len;int hash1, hash2;
} tr[N << 2];
void init()
{p[0] = 1;for (int i = 1; i <= n; i++)p[i] = p[i - 1] * P % mod;
}
int qpow(int a, int b)
{int res = 1;while (b){if (b & 1)res = res * a % mod;a = a * a % mod;b >>= 1;}return res;
}
void pushup(node &u, node &l, node &r)
{u.hash1 = (l.hash1 + r.hash1) % mod;u.hash2 = (l.hash2 + r.hash2) % mod;
}
void pushup(int u)
{pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void build(int u, int l, int r)
{tr[u] = {l, r, r - l + 1, 0, 0};if (l == r){tr[u] = {l, r, 1, (int)s[l] * p[n - l], (int)s[r] * p[r - 1]};return;}int mid = l + r >> 1;build(u << 1, l, mid);build(u << 1 | 1, mid + 1, r);pushup(u);
}
void modify(int u, int x, int k)
{if (tr[u].l == x && tr[u].r == x){tr[u].hash1 = k * p[n - x] % mod;tr[u].hash2 = k * p[x - 1] % mod;return;}int mid = tr[u].l + tr[u].r >> 1;if (x <= mid)modify(u << 1, x, k);elsemodify(u << 1 | 1, x, k);pushup(u);
}
int query1(int u, int l, int r)
{if (tr[u].l >= l && tr[u].r <= r)return tr[u].hash1;else{int res = 0;int mid = tr[u].l + tr[u].r >> 1;if (r > mid)res += query1(u << 1 | 1, l, r);if (l <= mid)res += query1(u << 1, l, r);return res % mod;}
}
int query2(int u, int l, int r)
{if (tr[u].l >= l && tr[u].r <= r)return tr[u].hash2;int res = 0;int mid = tr[u].l + tr[u].r >> 1;if (l <= mid)res += query2(u << 1, l, r);if (r > mid)res += query2(u << 1 | 1, l, r);return res % mod;
}
void solve()
{cin >> n >> q;cin >> (s + 1);init();build(1, 1, n);while (q--){int op;cin >> op;if (op == 2){int l, r;cin >> l >> r;int x = query1(1, l, r);int y = query2(1, l, r);x = (x * qpow(p[n - r], mod - 2)) % mod;y = (y * qpow(p[l - 1], mod - 2)) % mod;// cout << x << ' ' << y << endl;if (x == y)cout << "Yes" << endl;elsecout << "No" << endl;}else{int x;char k;cin >> x >> k;modify(1, x, (int)k);}}
}
signed main()
{Ysanqian;int T;T = 1;// cin >> T;while (T--)solve();return 0;
}