1 条题解
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#include <bits/stdc++.h> #define int long long using namespace std; int T; string s; int n; // s.length() // Hash const int MOD1 = 1000000007, MOD2 = 998244353; const int P1 = 29, P2 = 31; int hash1[30005], hash2[30005]; int powP1[30005], powP2[30005]; void initHash(int n) { powP1[0] = powP2[0] = 1; for (int i = 1; i <= n; i++) { hash1[i] = (hash1[i - 1] * P1 + s[i] - 'a' + 1) % MOD1; hash2[i] = (hash2[i - 1] * P2 + s[i] - 'a' + 1) % MOD2; powP1[i] = powP1[i - 1] * P1 % MOD1; powP2[i] = powP2[i - 1] * P2 % MOD2; } powP1[n + 1] = powP1[n] * P1 % MOD1; powP2[n + 1] = powP2[n] * P2 % MOD2; } bool hashCmp(int l, int r, int x, int y) { // return [l,r] == [x,y] int hash1LR = (hash1[r] - hash1[l - 1] * powP1[r - l + 1] % MOD1 + MOD1) % MOD1; int hash2LR = (hash2[r] - hash2[l - 1] * powP2[r - l + 1] % MOD2 + MOD2) % MOD2; int hash1XY = (hash1[y] - hash1[x - 1] * powP1[y - x + 1] % MOD1 + MOD1) % MOD1; int hash2XY = (hash2[y] - hash2[x - 1] * powP2[y - x + 1] % MOD2 + MOD2) % MOD2; return hash1LR == hash1XY && hash2LR == hash2XY; } //以 i 结尾有多少个 AA int f[30005]; //以 i 开头有多少个 AA int g[30005]; signed main() { // ios::sync_with_stdio(false); // cin.tie(0); cin >> T; while (T--) { cin >> s; n = s.length(); s = "^" + s + "$"; initHash(n); for (int i = 0; i <= n; i++) f[i] = g[i] = 0; // 处理 f // 枚举参考点间隙长度 L for (int L = 1; L < n; L++) { //枚举相邻参考点 for (int i = 1, j = i + L; j <= n; i += L, j += L) { int p, q; // i、j 前/后 最长公共长度 //二分计算 i,j 前面最长的公共 int l = 0; int r = i; p = 0; while (l <= r) { int mid = (l + r) / 2; // [i - mid + 1, i] == [j - mid + 1, j] if (hashCmp(i - mid + 1, i, j - mid + 1, j)) { p = mid; l = mid + 1; } else { r = mid - 1; } } l = 0; r = n - j; q = 0; while (l <= r) { int mid = (l + r) / 2; //[i + 1, i + mid] == [j + 1, j + mid] if (hashCmp(i + 1, i + mid, j + 1, j + mid)) { q = mid; l = mid + 1; } else { r = mid - 1; } } //长度为 L 的可行区间:[j+L-p,j+q] && [j,j+L-1] && [j,n] l = max(j + L - p, min(j, n)); r = min(j + q, min(j + L - 1, n)); // cout << L << ":" << i << "," << j << " " << p << "," << q << "\n"; if (l <= r) { // cout << l << "~" << r << endl; f[l]++, f[r + 1]--; } } } for (int i = 1; i <= n; i++) f[i] += f[i - 1]; // 处理 g // 枚举参考点间隙长度 L for (int L = 1; L < n; L++) { //枚举相邻参考点 for (int j = n, i = j - L; i >= 1; j -= L, i -= L) { int p, q; // i、j 前/后 最长公共长度 //二分计算 i,j 前面最长的公共 int l = 0; int r = i; p = 0; while (l <= r) { int mid = (l + r) / 2; // [i - mid + 1, i] == [j - mid + 1, j] if (hashCmp(i - mid + 1, i, j - mid + 1, j)) { p = mid; l = mid + 1; } else { r = mid - 1; } } l = 0; r = n - j; q = 0; while (l <= r) { int mid = (l + r) / 2; //[i + 1, i + mid] == [j + 1, j + mid] if (hashCmp(i + 1, i + mid, j + 1, j + mid)) { q = mid; l = mid + 1; } else { r = mid - 1; } } //长度为 L 的可行区间:[i-p+1,i-L+1+q] && [i-L+1,i] && [1,i-1] l = max(i - p + 1, max(i - L + 1, 1LL)); r = min(i - L + 1 + q, max(i, 1LL)); // cout << L << ":" << i << "," << j << " " << p << "," << q << "\n"; if (l <= r) { // cout << l << "~" << r << endl; g[l]++, g[r + 1]--; } } } for (int i = 1; i <= n; i++) g[i] += g[i - 1]; /* for (int i = 1; i <= n; i++) cout << f[i] << " "; cout << "\n"; for (int i = 1; i <= n; i++) cout << g[i] << " "; cout << "\n"; */ int ans = 0; // f[i],g[i+1] for (int i = 2; i <= n - 2; i++) ans += f[i] * g[i + 1]; cout << ans << "\n"; } return 0; }
- 1
信息
- ID
- 66
- 时间
- 1500ms
- 内存
- 512MiB
- 难度
- 9
- 标签
- 递交数
- 14
- 已通过
- 2
- 上传者