제출 #141442

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
141442		12tqian	Palindromic Partitions (CEOI17_palindromic)	C++14	100 / 100	437 ms	36724 KiB

palindromic

#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
const double PI = 4 * atan(1);
 
#define sz(x) (int)(x).size()
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair <int, int>
#define vi vector<int>
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define vpi vector<pair<int, int>>
#define vpd vector<pair<double, double>>
#define pd pair<double, double>
 
#define f0r(i,a) for(int i=0;i<a;i++)
#define f1r(i,a,b) for(int i=a;i<b;i++)
#define trav(a, x) for (auto& a : x)
 
void fast_io(){
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
}
void io(string taskname){
    string fin = taskname + ".in";
    string fout = taskname + ".out";
    const char* FIN = fin.c_str();
    const char* FOUT = fout.c_str();
    freopen(FIN, "r", stdin);
    freopen(FOUT, "w", stdout);
    fast_io();
}
 
const ll MOD = 1e9 +7;
const ll BASE = 131;
long long modInverse(long long b){
    long long ex = MOD - 2;
    if (b==1){
        return 1;
    }
    long long r = 1;
    while (ex ){
        if (ex&1){
            r=(r * b)%MOD;
        }
        ex = ex >> 1;
        b = (b * b)%MOD;
    }
    return r;
}
struct hsh {
    vector<ll> pre;
    int sz;
    ll INV;
    vector<ll> inv;
    vector<ll> po;
    void init(string s){
        sz = sz(s);
        pre.resize(sz);
        inv.resize(sz + 5);
        po.resize(sz + 5);
        INV = modInverse(BASE);
        f0r(i, sz(inv)){
            if(i == 0) inv[i] = 1;
            else inv[i] = (inv[i-1]*INV)%MOD;
        }
        f0r(i, sz(po)){
            if(i == 0) po[i] = 1;
            else po[i] = (po[i-1]*BASE)%MOD;
        }
        f0r(i, sz){
            if(i == 0) pre[i] =  (s[i] - 'a')*po[i];
            else pre[i] = pre[i-1] + (s[i] - 'a')*po[i];
        }
    }
    ll get(int l, int r){
        ll ret = (pre[r] - (l == 0? 0: pre[l-1]))%MOD;
        ret *= inv[l];
        ret %= MOD;
        return ret;
    }
};
 
 
int solve(string s){
    int n = sz(s);
    int lo = 0;
    int hi = n-1;
    int l = lo;
    int r = hi;
    int ret = 0;
    hsh h;
    h.init(s);
    //bonobo
    while(true){
    	if(lo>hi) break;
        if(l >= r){
            ret++;
            break;
        }
        if(h.get(lo, l) == h.get(r, hi)){
            ret += 2;
            lo = l;
            hi = r;
            lo++;
            hi--;
            l = lo;
            r = hi;
        }
        else{
            l++;
            r--;
        }
    }
    return ret;
}
int main(){
    fast_io();
    int t;
    cin >> t;
    string temporary;
    getline(cin, temporary);
    f0r(i, t){
        string s;
        getline(cin, s);
        cout << solve(s) << endl;
    }
    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

palindromic.cpp:1:0: warning: ignoring #pragma comment  [-Wunknown-pragmas]
 #pragma comment(linker, "/stack:200000000")
 
palindromic.cpp: In function 'void io(std::__cxx11::string)':
palindromic.cpp:47:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
     freopen(FIN, "r", stdin);
     ~~~~~~~^~~~~~~~~~~~~~~~~
palindromic.cpp:48:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
     freopen(FOUT, "w", stdout);
     ~~~~~~~^~~~~~~~~~~~~~~~~~~

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