Submission #1244747

#	Time	Username	Problem	Language	Result	Execution time	Memory
1244747		Minbaev	Mutating DNA (IOI21_dna)	C++20	0 / 100	24 ms	2368 KiB

dna

 #include "dna.h"
//#include "grader.cpp"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("Ofast,unroll-loops")

using namespace std;
using namespace __gnu_pbds;

#define pb      push_back
#define all(x)  x.begin(),x.end()
#define ar      array
#define mrand(a, b)   uniform_int_distribution<int>(a, b)(rng)

template<class T>bool umax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T>bool umin(T &a,T b){if(b<a){a=b;return true;}return false;}

template<class T> using ste = tree<T, null_type, less_equal<T>,
rb_tree_tag, tree_order_statistics_node_update>;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

namespace FAST {
    template<typename T, typename F>
    istream &operator>>(istream &cin, pair<T, F> &p) {
        cin >> p.first >> p.second;
        return cin;
    }

    template<typename T, typename F>
    ostream &operator<<(ostream &cout, pair<T, F> &p) {
        cout << p.first << ' ' << p.second;
        return cout;
    }

    template<typename T>
    istream &operator>>(istream &cin, vector<T> &a) {
    for (T &i: a) cin >> i;
        return cin;
    }

    template<typename T>
    ostream &operator<<(ostream &cout, vector<T> &a) {
          for (T i: a) cout << i << ' ';
        return cout;
    }

    template<typename T>
    istream &operator>>(istream &cin, deque<T> &a) {
        for (T &i: a) cin >> i;
        return cin;
    }

    template<typename T>
    ostream &operator<<(ostream &cout, deque<T> &a) {
        for (T i: a) cout << i << ' ';
        return cout;
    }
}
using namespace FAST;

const long long inf = 1e17 + 7;
const int mod = 1e9 + 7;
const int md = 998244353;

string A, B;
void init(string a, string b){
    A = a;
    B = b;
}

int get_distance(int x, int y) {

    vector<int>g[30], cnt(30);
    for(int i = x;i<=y;i++){
        g[B[i]-'A'].pb(i);
    }

    int res = 0;
    for(int i = x;i<=y;i++){
        if(g[A[i]-'A'].size() <= cnt[A[i]-'A']){
            return -1;
        }
        if(g[A[i]-'A'][cnt[A[i]-'A']] > i){
            res += g[A[i]-'A'][cnt[A[i]-'A']] - i;
        }
    }

    return res;

}

• Subtask 1: 0 / 21
• Subtask 2: 0 / 22
• Subtask 3: 0 / 13
• Subtask 4: 0 / 28
• Subtask 5: 0 / 16