Ural 1989 (tree-like array + polynomial hash)

Test instructions: Given a string, there are two operations, one is p a B, the string from position A to the substring of position B is a palindrome substring, the other operation c a B, the character of the string position A is replaced by B.

Solution: Because the string length is 1e5 and the number of times to ask also has 1e5, so the violence must be timed out, and then consider using a tree array to maintain the hash value of the string to solve, two operations with positive and negative direction of the interval ratio of the hash value and a single point modification.

#include <cstdio> #include <cstring> #include <algorithm> #define ULL unsigned long longusing namespace std;const int n = 100005;int n, m, L, R, Pos;char Str[n], q[20], C; ULL c[n][2], hash[n];int lowbit (int x) {return x & (-X);} ULL Sum (int x, int y) {ULL ret = 0;while (x > 0) {ret + c[x][y];x-= lowbit (x);} return ret;} void Add (int x, ULL d, int y) {while (x <= N) {c[x][y] + = d;x + = lowbit (x);}} int main () {hash[0] = 1;for (int i = 1; i < N; i++) hash[i] = hash[i-1] * 27;//are lowercase while (scanf ("%s", str) = = 1) {Me Mset (c, 0, sizeof (c)), n = strlen (str), for (int i = 0; i < n; i++) {ADD (i + 1, (Str[i]-' a ') * Hash[i], 0); ADD (i + 1, (Str[n-i-1]-' a ') * hash[i], 1);} scanf ("%d", &m), while (m--) {scanf ("%s", Q), if (q[0] = = ' P ') {scanf ("%d%d", &l, &r); ULL H1 = (sum (r, 0)-sum (l-1, 0)) * Hash[n-r]; ULL h2 = (SUM (n-l + 1, 1)-SUM (n-r, 1)) * HASH[L-1];IF (H1 = H2) printf ("yes\n"); elseprintf ("no\n");} else {scanf ("%d%c", &pos, &c);T x = c-str[pos-1]; ADD (POS, x * hash[pos-1], 0); ADD (N-pos + 1, x * Hash[n-pos], 1); str[pos-1] = C;}}} return 0;}

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Ural 1989 (tree-like array + polynomial hash)