Submission #939255

#	Time	Username	Problem	Language	Result	Execution time	Memory
939255		vjudge1	Palindromic Partitions (CEOI17_palindromic)	C++17	60 / 100	10004 ms	42428 KiB

palindromic

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
 
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define int long long
#define rnd(l, r) uniform_int_distribution<int>(l, r)(rng)
 
using namespace std;

void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);} 
int pow(int a,int b,int m){int ans=1;while(b){if(b&1){ans=(ans*a)%m;}b>>=1;a=(a*a)%m;}return ans;}
int binpow(int a,int b){int ans=1;while(b){if(b&1){ans=(ans*a);}b>>=1;a=(a*a);}return ans;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int mod = 1e8 + 7, N = 1e6 + 100, inf = 1e9, mod2 = 1e9 + 7;

int pw[N], hsh[N], hsh2[N], pw2[N];

int get(int l, int r){
  return (((hsh[r]-hsh[l-1]+mod)%mod)*pw[N - r])%mod; 
}
int get2(int l, int r){
	return (((hsh2[r]-hsh2[l-1]+mod2)%mod2)*pw2[N - r])%mod2;
}
void solve(){
	string s;
	cin >> s;
	int n = s.size();
	s = ' ' + s;
	
	for(int i= 1; i <= n;i++){
		hsh[i]=(hsh[i-1]+((s[i]-'a'+1)*pw[i-1])%mod)%mod;
		hsh2[i]=(hsh2[i-1]+((s[i]-'a'+1)*pw2[i-1])%mod2)%mod2;
	}
	
	vector <int> dp(n + 1, -inf);
	
	dp[0] = 0;
	int ans = 1;
	int mid = n / 2;
	for(int i = 1; i <= mid; i++){
		for(int j = i; j >= 1; j--){
			swap(i, j);
			if(get(n - j + 1,n - i + 1) == get(i, j) && get2(n - j + 1,n - i + 1) == get2(i, j)){
				dp[j] = max(dp[i - 1] + 2, dp[j]);
				ans = max(dp[j], ans);
				swap(i, j);
				break;
			}
			swap(i, j);
		}
	}
	
	for(int i = 0; i <= (n - 1) / 2; i++){
		ans = max(dp[i] + 1, ans);
	}
	
	cout << ans << endl;
}
 
main(){
	iostream::sync_with_stdio(false);	
    cin.tie(nullptr);
    cout.tie(nullptr);
    pw[0]=1;
    pw2[0] = 1;
	for(int i=1;i<N;i++){
		pw[i]=(pw[i-1]*27)%mod;
		pw2[i] = (pw2[i - 1] * 31) % mod2;
	}
	int t;
	cin >> t;
	while(t--)
		solve();
}	
/*
 * Before implementing the problem:
	
	Do I understand the problem correctly?
	Which places are tricky? What do I need to remember to implement them correctly?
	Which place is the heaviest by implementation? Can I do it simpler?
	Which place is the slowest? Where do I need to be careful about constant factors and where I can choose slower but simpler implementation?
	----------------------------------
	If not AC:
 
	Did you remember to do everything to handle the tricky places you thought about before?
	Is the solution correct?
	Do I understand the problem correctly?
	----------------------------------
	If you have a test on which the solution gives wrong answer:
 
	Is the solution doing what it was supposed to do?
	Is the problem in the code or in the idea?
*/

Compilation message (stderr)

palindromic.cpp:66:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   66 | main(){
      | ^~~~
palindromic.cpp: In function 'void fp(std::string)':
palindromic.cpp:15:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
palindromic.cpp:15:70: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                                                               ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 15 / 15
• Subtask 2: 20 / 20
• Subtask 3: 25 / 25
• Subtask 4: 0 / 40