SDUT F - 判断回文串

F - 判断回文串

用线段树去维护字符串前缀哈希，基础的字符串哈希是在一整段哈希中截取一部分哈希，但线段树的维护恰好将其反转过来，是维护区间哈希，通过一种 与字符串哈希获取某一段哈希 相反的方法将其维护上去。

这里贴一个写题的时候绷不住的笑话，N0和No是两个东西。Wa了N发才找到问题

ACcode 

#include 
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#define int long long
#define endl '\n'
#define lowbit(x) (x &-x)
#define mh(x) memset(x, -1, sizeof h)
#define debug(x) cerr << #x << "=" << x << endl;
#define brk exit(0);
using namespace std;
void TLE(){ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);}
const int N = 2e5 + 10;
const int M = 2 * N;
const int mod = 998244353;
const double esp = 1e-6;
const double pi = acos(-1);
typedef pair<int, int> PII;
typedef long long ll;
int pp[N];
struct node
{
    int hl, hr;
}t[N<<2];
string s;
void pushup(node& p, node &p1, node &p2, int  l, int r)
{
    
    p.hl = p1.hl + p2.hl * pp[l];
    p.hr = p1.hr * pp[r] + p2.hr;
}
 
void build(int p, int l, int r)
{
    if (l == r)
    {
        t[p].hl = t[p].hr = s[l];
        return;
    }
    int mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(t[p],t[p<<1],t[p<<1|1],mid-l+1,r-mid);
}
void modify(int p, int l, int r,int pos,char x)
{
    if (l == r)
    {
        t[p].hl = t[p].hr = x;
        return;
    }
    int mid = l + r >> 1;
    if (pos <= mid)modify(p << 1, l, mid, pos, x);
    else modify(p << 1 | 1, mid + 1, r, pos, x);
    pushup(t[p],t[p<<1],t[p<<1|1],mid-l+1,r-mid);
}
 
node query(int p, int l, int r, int ll, int rr)
{
    if (l >= ll && r <= rr)
    {
        return t[p];
    }
    int f1 = 0, f2 = 0;
    int mid = l + r >> 1;
    node res1, res2, ans;
    if (mid >= ll)res1 = query(p << 1, l, mid, ll, rr), f1 = 1;
    if (mid < rr)res2 = query(p << 1 | 1, mid + 1, r, ll, rr), f2 = 1;
    if (f1 && f2)
        pushup(ans, res1, res2, (mid - max(l, ll) + 1), (min(r, rr) - mid));
    else if (f1)
        ans= res1;
    else if (f2)
        ans = res2;
    return ans;
}
 
signed main()
{
    TLE();
    int n, m;
    cin >> n >> m;
    cin >> s;
    s = '0' + s;
    pp[0] = 1;
    for (int i = 1;i <= n;i++)
        pp[i] = pp[i - 1] * 137;
    build(1, 1, n);
    while(m--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int pos;char x;
            cin >> pos >> x;
            modify(1, 1, n, pos, x);
        }
        else if (op == 2)
        {
            int l, r;
            cin >> l >> r;
            node res = query(1, 1, n, l, r);
            if (res.hl == res.hr)
                cout << "Yes" << endl;
            else cout << "N0" << endl;
        }
    }
 
    return 0;
}