1902E - Collapsing Strings - CodeForces Solution

#include<bits/stdc++.h>
//#include<atcoder/all>
using namespace std;
//using namespace atcoder;
typedef int64_t ll;
//using mint=modint998244353;
typedef unsigned long ulong;

typedef long double ld;
const ll MODA=998244353;
ll vx[4]={0,1,0,-1};
ll vy[4]={1,0,-1,0};
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
long long gcd(long long a,long long b){
    a=abs(a);
    b=abs(b);
    ll gcdmax=max(a,b);
    ll gcdmin=min(a,b);
    if(gcdmin==0)return gcdmax;
    while(true){
        if(gcdmax%gcdmin==0)break;
        else gcdmax%=gcdmin;
        swap(gcdmin,gcdmax);
    }
    return gcdmin;
}
ll pow(ll N,ll P,ll M){
    if(P==0)return 1;
    else if(P%2==0){
        ll t=pow(N,P/2,M);
        return t*t%M;
    }
    else return N*pow(N,P-1,M)%M;
}
ll pow(ll N,ll P){
    if(P==0)return 1;
    else if(P%2==0){
        ll t=pow(N,P/2);
        return t*t;
    }
    else return N*pow(N,P-1);
}

vector<ll> fac;
vector<ll> finv;
vector<ll> inv;
void COMinit(ll N,ll P){
    rep(i,N+1){
        if(i==0){
            fac.push_back(1);
            finv.push_back(1);
            inv.push_back(1);
        }
        else if(i==1){
            fac.push_back(1);
            finv.push_back(1);
            inv.push_back(1);
        }
        else{
            fac.push_back(fac.at(i-1)*i%P);
            inv.push_back(P-inv.at(P%i)*(P/i)%P);
            finv.push_back(finv.at(i-1)*inv.at(i)%P);
        }
    }
}

ll COM(ll n,ll k,ll P){
    if(n<k)return 0;
    if(n<0||k<0)return 0;
    return fac.at(n)*(finv.at(k)*finv.at(n-k)%P)%P;
}


struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(ll N) : par(N) { //最初は全てが根であるとして初期化
        for(ll i = 0; i < N; i++) par[i] = i;
    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }
};

const ulong MASK30 = (1UL << 30) - 1;
const ulong MASK31 = (1UL << 31) - 1;
const ulong MOD = (1UL << 61) - 1;
const ulong MASK61 = MOD;

//mod 2^61-1を計算する関数
ulong CalcMod(ulong x)
{
    ulong xu = x >> 61;
    ulong xd = x & MASK61;
    ulong res = xu + xd;
    if (res >= MOD) res -= MOD;
    return res;
}

//a*b mod 2^61-1を返す関数(最後にModを取る)
ulong Mul(ulong a, ulong b)
{
    ulong au = a >> 31;
    ulong ad = a & MASK31;
    ulong bu = b >> 31;
    ulong bd = b & MASK31;
    ulong mid = ad * bu + au * bd;
    ulong midu = mid >> 30;
    ulong midd = mid & MASK30;
    return CalcMod(au * bu * 2 + midu + (midd << 31) + ad * bd);
}


int main(){
    ll N;
    cin>>N;
    vector<string> S(N);
    map<pair<ulong,ll>,int> mp;
    ll v=0;
    mt19937_64 rng(time(0));
    ulong base= rng() % MOD;
    ulong base2 =rng()%MODA;
    rep(i,N){
        cin>>S[i];
        v+=S[i].size();
        ulong hs=0;
        ulong hs2=0;
        rep(j,S[i].size()){
            ulong c=(S[i][j]-'a');
            c++; 
            hs=CalcMod(Mul(hs,base)+c);
            hs2=CalcMod(Mul(hs2,base2)+c);
            mp[{hs,hs2}]++;

        }

    }
    ll ans=2*N*v;
    rep(i,N)reverse(S[i].begin(),S[i].end());
    rep(i,N){
        cin>>S[i];
        v+=S[i].size();
        ulong hs=0;
        ulong hs2=0;
        rep(j,S[i].size()){
            ulong c=(S[i][j]-'a');
            c++; 
            hs=CalcMod(Mul(hs,base)+c);
            hs2=CalcMod(Mul(hs2,base2)+c);
            ans-=2*mp[{hs,hs2}];
        }
    }
    cout<<ans<<endl;
}