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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#596028 | #5065. Beautiful String | 333zhan | TL | 1ms | 3636kb | C++20 | 2.7kb | 2024-09-28 14:58:49 | 2024-09-28 14:58:50 |
Judging History
answer
#include <bits/stdc++.h>
#define int long long
using namespace std;
constexpr int P = 1145141;
constexpr int mod2 = 1E9 + 33;
constexpr int mod = 998244353;
constexpr int N = 5010;
constexpr int Z = 1E9 + 50;
int p[N];
int p2[N];
void solve () {
string s;
cin >> s;
const int n = s.size ();
s = " " + s;
vector <int> Hash (n + 1);
for (int i = 1; i <= n; i ++) {
Hash[i] = (Hash[i - 1] * P + s[i]) % mod;
}
vector <int> Hash2 (n + 1);
for (int i = 1; i <= n; i ++) {
Hash2[i] = (Hash2[i - 1] * P + s[i]) % mod2;
}
auto get = [&] (int l, int r) {
int len = r - l + 1;
int x = (Hash[r] - Hash[l - 1] * p[len] % mod + mod) % mod;
int y = (Hash2[r] - Hash2[l - 1] * p2[len] % mod2 + mod2) % mod2;
return x * Z + y;
};
vector <vector <int>> a (n + 1);
for (int i = 2; i <= n; i ++) {
for (int j = 1; j < i; j ++) {
if (i - j > n - i + 1) {
break;
}
int len = i - j;
if (get (j, i - 1) == get (i, i + len - 1)) {
a[i].push_back (j);
}
}
}
unordered_map <int, vector <pair <int, int>>> mp;
for (int l = 1; l <= n; l ++) {
for (int r = l; r <= n; r ++) {
mp[get (l, r)].push_back ({l, r});
}
}
int ans = 0;
for (auto &[_, v] : mp) {
const int m = v.size ();
for (int i = 0; i < m - 1; i ++) {
auto [l, r] = v[i];
if (r - l + 1 == 1) {
break;
}
int len = r - l + 1;
int x = lower_bound (a[l].begin (), a[l].end (), l - len + 1) - a[l].begin ();
int y = lower_bound (v.begin () + i, v.end (), pair <int, int> {r + 2, 0}) - v.begin ();
int sum2 = m - y;
// if (l == 2 && r == 3) {
// cerr << x << " " << sum2 << " " << m << '\n';
// }
if (v[i + 1].first == r + 1) {
sum2 --;
}
// if (l == 2 && r == 3) {
// cerr << sum2 << " " << a[l].size () - x << '\n';
// }
ans += sum2 * ((int)(a[l].size ()) - x);
}
}
cout << ans << '\n';
}
signed main () {
ios::sync_with_stdio (false);
cin.tie (nullptr);
p[0] = 1;
for (int i = 1; i < N; i ++) {
p[i] = p[i - 1] * P % mod;
}
p2[0] = 1;
for (int i = 1; i < N; i ++) {
p2[i] = p2[i - 1] * P % mod2;
}
int T = 1;
cin >> T;
while (T --) {
solve ();
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3636kb
input:
2 114514 0000000
output:
1 3
result:
ok 2 number(s): "1 3"
Test #2:
score: -100
Time Limit Exceeded
input:
11 79380 2483905227 37902399703986576270 39991723655814713848046032694137883815354365954917 5541883522667841718787744915558298207485830622522715211018760095628594390563630840464643840093757297 56530485554219245862513973750218715585679416120445975390556326891488719311495909340757506478400624741858999...