URAL-1989 树状数组+字符串哈希

Posted 2021-01-09 qaqorz

题意：给一个字符串s，两个操作，一个是询问s[l..r]是否回文，另一个是把s[i]的字符变成c

思路：判断回文可以做正反两个哈希，容易想到修改可以用树状数组维护，不过多项式就要反过来

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #define LL long long
 6 #define uLL unsigned long long
 7 #define debug(x) cout << "[" << x << "]" << endl
 8 using namespace std;
 9 
10 const int mx = 1e5+10;
11 const int base = 27;
12 uLL bit[mx][2], p[mx];
13 char s[mx];
14 int n;
15 
16 void upd(int x, int y, uLL c){
17     while (x <= n){
18         bit[x][y] += c;
19         x += x&-x;
20     }
21 }
22 
23 uLL sum(int x, int y){
24     uLL ans = 0;
25     while (x){
26         ans += bit[x][y];
27         x -= x&-x;
28     }
29     return ans;
30 }
31 
32 int main(){
33     int q;
34     p[0] = 1;
35     for (int i = 1; i < mx; i++) p[i] = p[i-1]*base;
36     while (scanf("%s", s) == 1){
37         n = strlen(s);
38         memset(bit, 0, sizeof bit);
39         for (int i = 0; i < n; i++){
40             upd(i+1, 0, p[i]*(s[i]-‘a‘));
41             upd(i+1, 1, p[i]*(s[n-i-1]-‘a‘));
42         }
43         scanf("%d", &q);
44         char op[20];
45         while (q--){
46             int x, y;
47             char c[2];
48             scanf("%s%d", op, &x);
49             if (op[0] == ‘p‘){
50                 scanf("%d", &y);
51                 uLL l = p[n-y]*(sum(y, 0)-sum(x-1, 0));
52                 uLL r = p[x-1]*(sum(n-x+1, 1)-sum(n-y, 1));
53                 printf("%s
", (l == r) ? "Yes" : "No");
54             }
55             else {
56                 scanf("%s", c);
57                 upd(x, 0, (c[0]-s[x-1])*p[x-1]);
58                 upd(n-x+1, 1, (c[0]-s[x-1])*p[n-x]);
59                 s[x-1] = c[0];
60             }
61         }
62     }
63     return 0;
64 }