Submission #1172940

#	Time	Username	Problem	Language	Result	Execution time	Memory
1172940		SpyrosAliv	Prisoner Challenge (IOI22_prison)	C++20	0 / 100	0 ms	328 KiB

prison

#include <bits/stdc++.h>
using namespace std;

int len(int x) {
    int sz = 0;
    while (x > 0) {
        x /= 10;
        sz++;
    }
    return sz;
}

int find_digit(int a, int d) {
    string s = "";
    while (a > 0) {
        s += ((a % 10) + '0');
        a /= 10;
    }
    reverse(s.begin(), s.end());
    for (int i = 0; i < 4; i++) s += '0';
    return s[d - 1] - '0';
}

string convert(int a) {
    string tr = "";
    while (a > 0) {
        tr += ((a % 3) + '0');
        a /= 3;
    }
    while (tr.size() < 7) tr += '0';
    reverse(tr.begin(), tr.end());
    return tr;
}

vector<vector<int>> devise_strategy(int n) {
    vector<vector<int>> s(28, vector<int>(n+1, -1));
    s[0][0] = 0;
    // the first prisoner will write the amount of digits A has
    // the second prisoner will compare this with the amount of digits B has
    // if they have same no. digits:
    /*
    the current prisoner will write the current digit of A on the board (form: pd, where p is the position of the digit and d is its value)
    the next prisoner will compare with the respective digit of B
    if they are equal, he will write which digit of A must be inspected next (added + 4) (so it doesnt conflict with the number written by the first pr.)
    
    max X: 49
    */

    /*
    Idea:
    We can convert the number to base 3 and compare them
    the prisoners will write on the board in the form vd, meaning that digit d has value of v

    even digits -> it corresponds to A
    odd digits  -> it corresponds to B
    */
    for (int i = 1; i <= n; i++) {
        string t = convert(i); // read A and write down the first digit of its trinary rep
        s[0][i] = (t[0] - '0') * 10 + 1;
    }
    for (int read = 1; read <= 27; read++) {
        int val = read / 10;
        int idx = read % 10 - 1; // which bit needs to be compared with val
        if (idx >= 8) continue;
        if (idx % 2 == 0) { // this corresponds to A
            s[read][0] = 1;
            for (int poss = 1; poss <= n; poss++) {
                string t = convert(poss);
                int k = t[idx] - '0';
                if (k < val) {
                    s[read][poss] = -2;
                }
                else if (k > val) {
                    s[read][poss] = -1;
                }
                else {
                    if (idx + 2 >= 8) s[read][poss] = -1;
                    else s[read][poss] = (t[idx + 1] - '0') * 10 + idx + 2;
                }
            }
        }
        else { // this corresponds to B
            s[read][0] = 0;
            for (int poss = 1; poss <= n; poss++) {
                string t = convert(poss);
                int k = t[idx] - '0';
                if (k > val) {
                    s[read][poss] = -2;
                }
                else if (k < val) {
                    s[read][poss] = -1;
                }
                else {
                    if (idx + 2 >= 8) s[read][poss] = -1;
                    else s[read][poss] = (t[idx + 1] - '0') * 10 + idx + 2;
                }
            }
        }
    }
    return s;
    /*
    for (int i = 1; i <= 4; i++) { // read i on the board
        s[i][0] = 1; // see B
        for (int j = 1; j <= n; j++) {
            if (len(j) < i) {
                s[i][j] = -2;
            }
            else if (len(j) > i) {
                s[i][j] = -1;
            }
            else s[i][j] = 1 + 4; // we need to see the first digit of A; add 1 so it doesnt conflict with the no. of digits of A
        }
    }
    for (int i = 5; i <= 8; i++) { // we need to see this digit of A next
        s[i][0] = 0;
        int need = i - 4;
        for (int j = 1; j <= n; j++) {
            s[i][j] = need * 10 + find_digit(j, need);
        }   
    }
    for (int i = 10; i <= 49; i++) { // we know that the i[0] digit of A is i[1]
        s[i][0] = 1;
        int v = i % 10;
        int d = i / 10;
        for (int j = 1; j <= n; j++) {
            int rep = find_digit(j, d);
            if (rep < v) {
                s[i][j] = -2;
            }
            else if (rep > v) {
                s[i][j] = -1;
            }
            else {
                s[i][j] = d + 1 + 4; // we must see the next digit
            }
        }
    }*/
    return s;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 90