oj.uz

Submission #357567

# Submission time Handle Problem Language Result Execution time Memory
357567 2021-01-24T06:28:00 Z blue PIN (CEOI10_pin) C++17
60 / 100
 47 ms 10348 KB 
#include <iostream>
#include <vector>
using namespace std;

/*
There are 36^4 = 1679616 possibilities for the PIN.

Let them be numbered from 0 to 1679615.
Two PINS have one difference if their
*/

int adj(char c)
{
    if(c == '0') return 1;
    else if('1' <= c && c <= '9') return (c - '1') + 2;
    else return (c - 'a') + 11;
}

int ct[37][37][37][37];

int main()
{
    int N, D;
    cin >> N >> D;

    string S;
    vector< vector<int> > T(N, vector<int>(4));
    for(int i = 0; i < N; i++)
    {
        cin >> S;
        for(int j = 0; j < 4; j++) T[i][j] = adj(S[j]);
        for(int mask = 0; mask < (1 << 4); mask++)
        {
            ct[(mask & (1 << 0)) ? T[i][0] : 0]
              [(mask & (1 << 1)) ? T[i][1] : 0]
              [(mask & (1 << 2)) ? T[i][2] : 0]
              [(mask & (1 << 3)) ? T[i][3] : 0]++;
        }
    }

    vector<long long> res(5, 0);

    vector<int> X;




    for(int i = 0; i < N; i++)
    {
        X = T[i];
        res[1] += ct[0][X[1]][X[2]][X[3]];
        res[1] += ct[X[0]][0][X[2]][X[3]];
        res[1] += ct[X[0]][X[1]][0][X[3]];
        res[1] += ct[X[0]][X[1]][X[2]][0];
        res[1] -= 4 * ct[X[0]][X[1]][X[2]][X[3]];
    }




    for(int i = 0; i < N; i++)
    {
        X = T[i];
        res[2] += ct[0][0][X[2]][X[3]];
        res[2] += ct[0][X[1]][0][X[3]];
        res[2] += ct[0][X[1]][X[2]][0];
        res[2] += ct[X[0]][0][0][X[3]];
        res[2] += ct[X[0]][0][X[2]][0];
        res[2] += ct[X[0]][X[1]][0][0];


        res[2] -= 6 * ct[X[0]][X[1]][X[2]][X[3]];
    }
    res[2] -= 3 * res[1];





    for(int i = 0; i < N; i++)
    {
        X = T[i];
        res[3] += ct[0][0][0][X[3]];
        res[3] += ct[0][0][X[2]][0];
        res[3] += ct[0][X[1]][0][0];
        res[3] += ct[X[0]][0][0][0];

        res[3] -= 4 * ct[X[0]][X[1]][X[2]][X[3]];
    }
    res[3] -= 3 * res[1];
    res[3] -= 2 * res[2];





    for(int i = 0; i < N; i++)
    {
        X = T[i];
        res[4] += ct[0][0][0][0];

        res[3] -= ct[X[0]][X[1]][X[2]][X[3]];
    }
    res[4] -= res[1];
    res[4] -= res[2];
    res[4] -= res[3];





    res[D] /= 2;
    cout << res[D] << '\n';
}

Subtask #1
60.0 / 100.0

# Verdict Execution time Memory Grader output
1 Correct 2 ms 2028 KB Output is correct
2 Correct 2 ms 2028 KB Output is correct
3 Incorrect 2 ms 1900 KB Output isn't correct
4 Correct 11 ms 2940 KB Output is correct
5 Correct 13 ms 3180 KB Output is correct
6 Correct 13 ms 3180 KB Output is correct
7 Correct 11 ms 2924 KB Output is correct
8 Correct 15 ms 3436 KB Output is correct
9 Correct 25 ms 4332 KB Output is correct
10 Correct 25 ms 4588 KB Output is correct
11 Correct 14 ms 3308 KB Output is correct
12 Incorrect 30 ms 4460 KB Output isn't correct
13 Incorrect 16 ms 3468 KB Output isn't correct
14 Incorrect 15 ms 3424 KB Output isn't correct
15 Incorrect 24 ms 4460 KB Output isn't correct
16 Correct 35 ms 9452 KB Output is correct
17 Correct 47 ms 10348 KB Output is correct
18 Incorrect 37 ms 9776 KB Output isn't correct
19 Incorrect 39 ms 10092 KB Output isn't correct
20 Incorrect 46 ms 10348 KB Output isn't correct