題目鏈接

題意

$N$次操作每次選擇在字符串$S$的開始或者末尾拆入一個字符串，每次操作詢問當前字符串一共有多少種循環節

思路

對最終的字符串做哈希處理，記錄每次操作對應字符串$S$的區間$[L,R]$，二分找到循環節長度為$X$的字符串的最后出現的操作$N_2$，長度X的循環節的貢獻$[N_1,N_2]$，最后統計答案即可

#include <bits/stdc++.h> #define endl '\n' const int maxn = 2e6 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; char s[10], t[maxn], ss[10]; int L[maxn], R[maxn]; int a[maxn]; struct Hash{long long p[maxn], hash[maxn], base = 131;long long getHash(int l, int r) {long long ans = (hash[r] - hash[l-1] * p[r-l+1]) % mod;return (ans + mod) % mod;}void init(char* buf) {int n = strlen(buf);p[0] = 1;for (int i = 1; i <= n; ++i) p[i] = p[i - 1] * base % mod;for (int i = 1; i <= n; ++i) {hash[i] = (hash[i - 1] * base % mod + (buf[i-1] - 'a' + 1)) % mod;}} }hh; int check(int q, int x) {return (hh.getHash(L[q], R[q]-x) == hh.getHash(L[q]+x, R[q])); } int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n, head = 1e6+1, tail = 1e6+2;scanf("%d", &n);for (int i = 1; i <= n; ++i) {scanf("%s%s", ss, s);char c = s[0];if (c == 's' && s[1] == 'i') c = 'z'; if (ss[0] == 'a') t[tail++] = c;else t[head--] = c;L[i] = head+1, R[i] = tail-1;}for (int i = 1; i <= n; ++i) L[i] -= head, R[i] -= head;t[tail] = 0;hh.init(t+head+1);for (int i = 1; i <= n; ++i) {int l = i, r = n;while (l <= r) {int mid = (l + r) >> 1;if (check(mid, i)) l = mid + 1;else r = mid - 1;}a[i]++, a[r+1]--;}for (int i = 1; i <= n; ++i) {a[i] += a[i-1];printf("%d\n", a[i]);}return 0; }

總結

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