Submission #1203192

#	Time	Username	Problem	Language	Result	Execution time	Memory
1203192		ender_shayan	Palindromic Partitions (CEOI17_palindromic)	C++20	15 / 100	10 ms	23948 KiB

palindromic

#include <bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

typedef long long ll;
typedef long double	ld;
typedef pair<int, int>	pii  ;
typedef pair<ll, ll>	pll  ;
typedef vector<pii>     vii  ;
typedef vector<int>     veci ;
typedef vector<pll>     vll  ;
typedef vector<ll>      vecll;

// find_by_order             order_of_key

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define F		        first
#define S		        second
#define pb		        push_back
#define endl            '\n'
#define Mp		        make_pair
#define all(x)          x.begin(), x.end()
#define debug(x)        cerr << #x << " = " << x << endl
#define set_dec(x)	    cout << fixed << setprecision(x);
#define fast_io         ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io         freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define lb              lower_bound
#define ub              upper_bound
#define for1(n)         for(int i=1;i<=n;i++)
#define for0(n)         for(int i=0;i<n;i++)
#define forn(n)         for(int i=n;i>0;i--)
#define pq              priority_queue <pii, vector<pii>, greater<pii>>


ll mod = 1e9+7;

ll inf=1e18;
const int N=1e6+100,L=21,bs=1000033,bs2=1000039;
int A[N],B[N],C[N],D[N],E[N],n,m,k,q,dp[N],pre[N],dist[N],vis[N];
vector<int>g[N];

ll pw[N],pw2[N];
int main(){
    fast_io
    int T;cin>>T;
    pw[0]=pw2[0]=1;
    for1(200)pw[i]=pw[i-1]*bs%mod;
    while(T--){
        string s;cin>>s;s='0'+s;
        n=s.size()-1;
        int l=1,r=n,ans=0;
        ll h1=0,h2=0,h3=0,h4=0;
        while(l<=n/2){
            h1=(h1+pw[s[l]])%mod;
            h2=(h2+pw[s[r]])%mod;
            h3=(h3+pw2[s[l]])%mod;
            h4=(h4+pw2[s[r]])%mod;
            if(h1==h2 && h3==h4)
                ans+=2;
            l++;r--;
        }
        ans+=(h1!=h2 || h3!=h4 || n%2==1);
        cout<<ans<<endl;
    }




}

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