結果
問題 | No.2204 Palindrome Splitting (No Rearrangement ver.) |
ユーザー |
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提出日時 | 2023-02-03 21:51:43 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 215 ms / 2,000 ms |
コード長 | 2,169 bytes |
コンパイル時間 | 3,982 ms |
コンパイル使用メモリ | 258,024 KB |
最終ジャッジ日時 | 2025-02-10 09:08:27 |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 33 |
ソースコード
#include<bits/stdc++.h>#include<atcoder/all>using namespace std;using namespace atcoder;typedef modint998244353 mint;typedef long long ll;// isprimebool isprime(ll n){if (n == 1) return false;for (ll i = 2; i * i <= n; i++){if (n % i == 0) return false;}return true;}// importrandom// don't forget randomjumon// and also isprimell RandomMod(ll l, ll r){ll ret = l + rand() % (r - l);while (!isprime(ret)) ret = l + rand() % (r - l);return ret;}ll randrange(ll l, ll r){return l + rand() % (r - l);}// -----int main(){ios_base::sync_with_stdio(false);cin.tie(NULL);srand((unsigned)time(NULL));// Rolling Hash// don't forget randomjumon and importrandomtypedef dynamic_modint<0> mint1;typedef dynamic_modint<1> mint2;int mod1 = RandomMod(7e8, 1e9);int mod2 = RandomMod(7e8, 1e9);mint1::set_mod(mod1);mint2::set_mod(mod2);mint1 b1 = randrange(100, 200);mint2 b2 = randrange(100, 200);int MAX_m = 10000;vector<mint1> b1l(MAX_m+1);b1l[0] = 1;for (int i=0; i<MAX_m; i++){b1l[i+1] = b1l[i] * b1;}vector<mint2> b2l(MAX_m+1);b2l[0] = 1;for (int i=0; i<MAX_m; i++){b2l[i+1] = b2l[i] * b2;}// -----int INF = 1e9;string s; cin >> s;int n = s.size();vector<vector<int>> dp(n+1, vector<int>(n+1, -INF));dp[0][0] = INF;mint1 r1 = 0;mint2 r2 = 0;vector<mint1> s1(n+1);vector<mint2> s2(n+1);vector<mint1> t1(n+1);vector<mint2> t2(n+1);for (int i=0; i<n; i++){r1 = r1 * b1 + s[i] - 'a' + 1;r2 = r2 * b2 + s[i] - 'a' + 1;s1[i+1] = r1;s2[i+1] = r2;}r1 = 0;r2 = 0;for (int i=n-1; i>=0; i--){r1 = r1 * b1 + s[i] - 'a' + 1;r2 = r2 * b2 + s[i] - 'a' + 1;t1[i] = r1;t2[i] = r2;}mint1 x1, y1;mint2 x2, y2;for (int i=0; i<n; i++){for (int j=0; j<=i; j++){x1 = s1[i+1] - s1[j] * b1l[i-j+1];x2 = s2[i+1] - s2[j] * b2l[i-j+1];y1 = t1[j] - t1[i+1] * b1l[i-j+1];y2 = t2[j] - t2[i+1] * b2l[i-j+1];if (x1 == y1 && x2 == y2){dp[i+1][i+1] = max(dp[i+1][i+1], min(dp[i][j], i-j+1));}}for (int j=0; j<=i; j++){dp[i+1][j] = max(dp[i+1][j], dp[i][j]);}}cout << dp[n][n] << endl;}