제출 #469298

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
469298		Carmel_Ab1	Cubeword (CEOI19_cubeword)	C++17	50 / 100	161 ms	26048 KiB

cubeword

#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld, ld> pld;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T>
ostream &operator<<(ostream &os, vector<T> &a) {
    os << "[";
    for (int i = 0; i < ll(a.size()); i++) { os << a[i] << ((i != ll(a.size() - 1) ? " " : "")); }
    os << "]\n";
    return os;
}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << "" << p.first << " " << p.second << "";
    return os;
}

void usaco(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);
}

template<typename T>
void read(vector<T> &v) {
    int n = v.size();
    for (int i = 0; i < n; i++)
        cin >> v[i];
}

template<typename T>
vector<T> UNQ(vector<T> a) {
    sort(all(a));
    vector<T> ans;
    for (T t:a)
        if (ans.empty() || t != ans.back())
            ans.push_back(t);
    return ans;
}

ll ceil(ll a, ll b) {
    ll ans = a / b;
    if (a % b)ans++;
    return ans;
}

ld log(ld a, ld b) { return log(b) / log(a); }

ll power(ll base, ll exp, ll M = 1e18) {//(base^exp)%M
    ll ans = 1;
    while (exp) {
        if (exp % 2 == 1)ans = ((ans % M) * (base % M)) % M;
        base = ((base % M) * (base % M)) % M;
        exp /= 2;
    }
    return ans;
}

string to_base(int n, int new_base) {
    string s;
    int nn = n;
    while (nn) {
        s += to_string(nn % new_base);
        nn /= new_base;
    }
    reverse(all(s));
    return s;
}

ll gcd(ll a, ll b) {
    if (a < b)swap(a, b);
    if (b == 0)return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    ll x = (a / gcd(a, b));
    return b * x;
}

vl generate_array(ll n, ll mn = 1, ll mx = 1e18 + 1) {
    if (mx == 1e18 + 1)
        mx = n;
    vl ans(n);
    for (int i = 0; i < n; i++)
        ans[i] = (mn + rand() % (mx - mn + 1));
    return ans;
}

string substr(string s, int l, int r) {
    string ans;
    for (int i = l; i <= r; i++)
        ans += s[i];
    return ans;
}


void solve();

int main() {
    GLHF;
    int t = 1;
    //cin >> t;
    while (t--)
        solve();
}

void solve() {
    int n;
    cin >> n;
    vector<string>a(n);

    read(a);
    for(int i=0; i<n; i++){
        reverse(all(a[i]));
        a.pb(a[i]);
        reverse(all(a[i]));
    }

    a=UNQ(a);
    vector<vector<vvl>>cnt(26,vector<vvl>(26,vvl(26,vl(11))));
    n=a.size();

    vector<vvl>cnt2(26,vvl(26,vl(11)));
    for(int i=0; i<n; i++)
        cnt2[a[i][0]-'a'][a[i].back()-'a'][a[i].size()]++;

    ll ans=0,mod=998244353;
    string AB="abcdefghijklmnopqrst";
    for(int l=1; l<=10; l++)
        for(char c1:AB)
            for(char c2:AB)
                for(char c3:AB)
                    for(char c4:AB){
                        ll cur=1;
                        cur=(cur*cnt2[c1-'a'][c4-'a'][l])%mod;
                        cur=(cur*cnt2[c2-'a'][c4-'a'][l])%mod;
                        cur=(cur*cnt2[c3-'a'][c4-'a'][l])%mod;

                        cnt[c1-'a'][c2-'a'][c3-'a'][l]+=cur;
                    }

    for(int l=1; l<=10; l++) {
        for (char c1:AB)
            for (char c2:AB)
                for (char c3:AB)
                    for (char c4:AB) {
                        ll cur = 1;
                        cur = (cur * cnt[c1 - 'a'][c2 - 'a'][c3 - 'a'][l]) % mod;
                        cur = (cur * cnt[c1 - 'a'][c3 - 'a'][c4 - 'a'][l]) % mod;
                        cur = (cur * cnt[c2 - 'a'][c3 - 'a'][c4 - 'a'][l]) % mod;
                        cur = (cur * cnt[c1 - 'a'][c2 - 'a'][c4 - 'a'][l]) % mod;
                        if(cur){
                            ll x=1;
                        }
                        ans = (ans + cur) % mod;
                    }
    }
    out(ans)
}

컴파일 시 표준 에러 (stderr) 메시지

cubeword.cpp: In function 'void solve()':
cubeword.cpp:190:32: warning: unused variable 'x' [-Wunused-variable]
  190 |                             ll x=1;
      |                                ^
cubeword.cpp: In function 'void usaco(std::string)':
cubeword.cpp:56:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   56 |     freopen(FIN, "r", stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~
cubeword.cpp:57:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   57 |     freopen(FOUT, "w", stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~

• Subtask 1: 21 / 21
• Subtask 2: 29 / 29
• Subtask 3: 0 / 34
• Subtask 4: 0 / 16